# Introduction to Pre-Calculus

These are my notes on an Introduction to Pre-calculus.

### Table of Contents

- Functions and Notation
- Domain and Range
- Rates of Change
- Composition of Functions
- Transformations of Functions
- Quadratic Models
- Inequalities with Quadratic Functions
- Polynomial Functions and Models
- Properties of Rational Functions
- Graphs of Rational Functions
- Polynomial Inequalities
- Real Zeros of Polynomials
- Complex Zeros

**Functions and Notation**

**Intro**

A relation is a set of ordered pairs. The set of the first components of each ordered pair is called the range. consider the following set of ordered pairs.

The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.

\(1,2 2,4 3,6 4,8 5,10\)

The domain is (1,2,3,4,5)

The range is (2,4,6,8,10)

Note that each value in the domain is also known as an input or independent

variable. It is often labeled with a lowercase x. Each value in the range is

also known as an output value or dependent variable. It is often labeled with

a lowercase y. **Functions**

A function f is a relation that assigns a single value in the range to each

value in the domain. In other words, no x-values are repeated. For our example

that relates the first five natural numbers to numbers double their value, this

relation is a function because each element in the domain (1,2,3,4,5) is

paired with exactly one element in the range (2,4,6,8,10).

Now lets consider the set of ordered pairs that relates the terms even

and odd to the first five natural numbers.

(odd,1) (even,2) (odd,3) (even,4) (odd,5)

Notice that each element in the domain (even,odd) is not paired with exactly

one element in the range, (1,2,3,4,5). For example, the term odd corresponds

to three values from the range (1,3,5) and the term even corresponds to two

values from the range (2,4). This violates the definition of a function,

so this relation is not a function.

A function is a relation in which each possible input value leads to exactly

one output value. We say the output is a function of the input. The input

values make up the domain, and the output values make up the range.

To decide if a relation is a function. identify the input values, identify

the output values. then: if each input value leads to only one output value,

classify the relationship as a function. If any input value leads to two or

more outputs, do not classify the relationship as a function.**Function Notation**

Once that we determine that a relationship is a function, we need to

display and define the functional relationship so that we can understand and use

them, and sometimes also so that we can program them into computers. There are

various ways of representing functions. A standard function notation is one

representation that facilitates working with functions.

To represent height is a function of age, we start by identifying the

descriptive variables "h" for height and "a" for age. The letters f,g, and h,

are often used to represent functions.

h is f of a - We name the function f, height is a function of age.

h=f(a) - We use parentheses to indicate the function input.

f(a) - We name the function f, the expression is read as "f of a".

Remember that we can use any letter to name the function. The notation

h(a) shows us that h depends on a. The value a must be put into the function h

to get a result. The parentheses indicate that age is input into the function,

they do not indicate multiplication.

We can also give an algebraic expression as the input to a function. For

example, f(a+b) means first a and b, and the result is the input for the

function f. The operations must be performed in this order to obtain the

correct result.

The notation y=f(x) defines a function named f. This is read as "y is a

function of x". The letter x represents the input value or independent

variable. The letter y or f(x) represents the output value or dependent

variable.**Example 1**

Determine if menu price lists are functions.

Is the price a function of the item?

Is the item a function of the price?

Solution:

Let's begin by considering the input as the items on the menu. The

output values are then the prices. Each item on the menu has only one price,

so the price is a function of the item.

Two items on the menu have the same price. If we consider the prices to be

the input values and the items to be the output, then the same input value

could have more than one output associated with it. So, the item is not a

function of the price.**Example 2**

Determine if class grade rules are functions

In a particular math class, the overall percent grade corresponds to a grade

point average. Is grade point average a function of the percent grade? Is the

percent grade a function of the grade point average?

Solution:

For any percent grade earned, there is an associated grade point average, so

the grade point average is a function of the percent grade. In other words, if

we input the percent grade, the output is a specific grade point average.

In the grading system given, there is a range of percent grades that

correspond to the same grade point average. For example, students who

receive a grade point average of 3.0 could have a variety of percent grades

ranging from 78 all the way to 86. Thus, the percent grade is not a function

of grade point average.**Example 3**

Using function notation for days in a month

Use function notation to represent a function whose input is the name of a

month and output is the number of days in that month. Assume that the domain

does not include leap years.

Solution:

The number of days in a month is a function of the name of the month, so if

we name the function f, we write days=f(month) or d-f(m). The name of the

month is the input to a rule that associates a specific number(the

output) with each input. For example, f(march)=31, because march has 31

days. The notation d-f(m) reminds us that the number of days, d(output), is

dependent on the name of the month, m(input).

Note that the inputs to a function do not have to be numbers. Function inputs

can be the names of people, labels of geometric objects, or any other

element that determines some kind of output. However, most of the functions we

will work with in this book will have numbers as inputs and outputs. **Example 4**

Interpreting function notation

A function N=f(y) gives the number of police officers, N, in a town in year y.

What does f(2005)=300 represent?

Solution:

When we read f(2005)=300, we see that the input year is 2005. The value for

the output, the number of police officers(N), is 300. Remember N=f(y). The

statement f(2005)=300 tells us that in the year 2005 there were 300 police

officers in town.**Representing Functions Using Tables**

A common method of representing functions is in the form of a table. The

table rows or columns display the corresponding input and output values. in

some cases, these values represent all we know about the relationship.; other

times, the table provides a few select examples from a more complete

relationship.**Finding Input and Output Values of a Function**

When we know an input value and want to determine the corresponding output

value for a function, we evaluate the function. Evaluating will always

produce one result because each input of a function corresponds to exactly

one output value.

When we know an output value and want to determine the input values that would

produce that output value, we set the output equal to the function's formula

and solve for the input. Solving can produce more than one solution because

different input values can produce the same output value.

When we have a function in formula form, it is usually a simple matter to

evaluate the function. For example, the function \(f(x)=5-3x^2\) can be

evaluated by squaring the input value, multiplying by 3, and then subtracting

the product from 5.**Example 6**

Evaluate \(f(x)=x^2+3x-4\) for x=2

Solution:

\(f(x)=(2)^2 + 3(2) - 4\)

\(f(x)=4+6-4\)

\(f(x)=6\)**Example 7**

Given the function \(h(p)=p^2+2p\) evaluate h(4)

Solution:

To evaluate h(4), we substitute the value 4 for the input variable p in the

given function.

\(h(p)=p^2+2p\)

\(h(4)=4^2+2(4)\)

\(=16+8\)

\(=24\)**Example 8**

Given the function \(h(p)=p^2+2p\) solve for h(p)=3

Solution:

\(h(p)=3\)

\(p^2+2p=3\)

\(p^2+2p-3=0\)

\((p+3)(p-1)=0\)

p=-3,1**Evaluating Function Expressed in Formulas**

Some functions are defined by mathematical rules or procedures expressed in

equation form. If it is possible to express the function output with a

formula involving the input quantity, then we can define a function in

algebraic form. For example, the equation \(2n+6p=12\) expresses a functional

relationship between n and p. We can rewrite it to decide if p is a function

of n.

Given a function in equation form, write its algebraic formula:

1. Solve the equation to isolate the output variable on one side of the equal

sign, with the other side as an expression that involves only the input

variable.

2. Use all the usual algebraic methods for solving equations, such as adding

or subtracting the same quantity to or from both sides, or multiplying or

dividing both sides of the equation by the same quantity. **Example 9**

Finding the Algebraic form of a Function

Express the relationship \(2n+6p=12\) as a function \(p=f(n)\) if possible.

Solution:

To express the relationship in this form, we need to be able to write the

relationship where p is a function of n, which means writing it as

p=[expression involving n].

\(2n+6p=12\)

\(6p=12-2n\)

\(p=\frac{12-2n}{6}\)

\(p=\frac{12}{6} - \frac{2n}{6}\)

\(p=2-\frac{1}{3}n\)

It is important to note that not every relationship expressed by an equation

can also be expressed as a function with a formula.**Example 10**

Expressing the Equation of a Circle as a Function

does the equation \(x^2+y^2=1\) represent a function with x as input and y as

output? If so, express the relationship as a function \(y=f(x)\).

Solution:

First we subtract \(x^2\) from both sides.

\(y^2=1-x^2\)

Now we try to solve for y in this equation

\(y=\pm \sqrt{1-x^2}\)

\(y=+ \sqrt{1-x^2} \text{ and} - \sqrt{1-x^2}\)

We get two outputs corresponding to the same input, so this relationship cannot

be represented as a single function \(y=f(x)\)

Are there relationships expressed by an equation that do represent a function

but which still cannot be represented by an algebraic formula?

Yes this can happen. For example, given the equation \(x=y+2^y\), if we want

to express y as a function of x, there is no simple algebraic formula

involving only x that equals y. However, each x does determine a unique value

for y, and there are mathematical procedures by which y can be found to any

desired accuracy. In this case, we say that the equation gives an

implicit rule for y as a function of x, even though the formula cannot be

written explicitly.**Evaluating a Function Given in Tabular Form**

As we saw above, we can represent functions in tables. Conversely, we can use

information in tables to write functions, and we can evaluate functions using

the tables.

Given a function represented by a table, identify specific output and input

values:

1. Find the given input in the row of input values

2. Identify the corresponding output value paired with that input value

3. Find the given output values in the row or column of output values, noting

every time that output value appears.

4. Identify the input value or values corresponding to the given output value.**Example 11**

Evaluating and Solving a Tabular Function

1. Evaluate g(3)

2. Solve \(g(n)=6\)

Solution:

Evaluating g(3) means determining the output value of the function g for the

input value of n=3. The table output value corresponding to n=3 is

7, so g(3)=7.

Solving g(n)=6 means identifying the input values n, that produce an output

value of 6. When we input 2 into the function g, our output is 6. When we

input 4 into the function g, our output is also 6.**Finding Function Values From a Graph**

Evaluating a function using a graph requires finding the corresponding value

for a given input value, only in this case, we find the output value by

looking at the graph. Solving a function equation using a graph requires

finding all instances of the given output value on the graph and observing the

corresponding input values.

1. Evaluate f(2)

2. Solve f(x)=4

Solution:

1. To evaluate f(2), locate the point on the curve where x=2, then read the

y-coordinate of that point. The point has coordinates of (2,1) so f(2)=1.

2. To solve f(x)=4, we find the output value 4 on the vertical axis. moving

horizontally along the line y=4, we locate two points of the curve with output

value 4:(-1,4) and (3,4). These points represent the two solutions to f(x)=4:

-1 or 3. This means f(-1)=4 and f(3)=4, or when the input is -1 or 3, the

output is 4.**Determining Whether a Function is One-to-One**

Some functions have a given output value that corresponds to two or

more input values. For example, in the stock chart shown, the stock price

was $1000 on five different dates, meaning that there were five different input

values that all resulted in the same output value of $1000.

However, some functions have only one input value for each output value, as

well as having only one output for each input. We call these functions

one-to-one functions. As an example, consider a school that uses only letter

grades and decimal equivalents.

This grading system represents a one to one function, because each letter input

yields one particular grade point average output and each grade point average

corresponds to one input letter.

To visualize this concept, let's look again at the two simple functions

sketched. The function in part 1 shows a relationship that is not a one to one

function because inputs q and r both give output n. The function in part 2

shows a relationship that is a one to one function because each input is

associated with a single output.

A one to one function is a function in which each output value corresponds

to exactly one input value.**Example 13**

Determining Whether a Relationship is a One to One function

Is the area of a circle a function of its radius? if yes, is the function one

to one?

Solution:

A circle of radius r has a unique area measure given by \(A=\pi r^2\) for any

input r, there is only one output. The area is a function of radius r.

If the function is one to one, the output value, the area, must correspond to

a unique input value, the radius. Any area measure A is given by the formula

\(A=\pi r^2\). Because areas and radii are positive numbers, there is exactly

one solution: \(\sqrt{A}{\pi}\). So the area of a circle is a one to one

function of the circle's radius. **Using the Vertical Line Test**

As we have seen in some examples above, we can represent a function

using a graph. Graphs display a great many input-output pairs in a small

space. The visual information they provide often makes relationships easier to

understand. by convention, graphs are typically constructed with the

input values along the horizontal axis and the output values along the vertical

axis.

The most common graphs name the input value y, and we say y is a function of x,

or y=f(x) when the function is named f. The graph of the function is the set

of all points (x,y) in the plane that satisfies the equation y=f(x). If the

function is defined for only a few input values, then the graph of the function

is only a few points, where the x-coordinate of each point is an input value

and the y-coordinate of each point is the corresponding output

value. for example, the black dots on the graph tell us that f(0)=2 and

f(6)=1. However, the set of all points (x,y) satisfying y=f(x) is a

curve. The curve includes (0,2) and (6,1) because the curve passes through

those points.

The vertical line test can be used to determine whether a graph

represents a function. If we can draw any vertical line that intersects a

graph more than once, then the graph does not define a function because a

function has only one output value for each input value.

Given a graph, use the vertical line test to determine if the graph represents a

function.

1. Inspect the graph to see if any vertical line drawn would intersect the curve

more than once.

2. If there is any such line, determine that the graph does not represent a

function.**Example 14**

Applying the vertical line test

Which of the graphs represent a function y=f(x)?

Solution:

If any vertical line intersects a graph more than once, the relation

represented by the graph is not a function. Notice that any vertical line

would pass through only one point of the two graphs shown in parts a and b.

From this we can conclude that these two graphs represent functions. The

third graph does not represent a function because, at most x-values, a vertical

line would intersect the graph at more than one point.**Using the Horizontal line test**

Once we have determined that a graph defines a function, an easy way to

determine if it is a one to one function is to use the horizontal line test. Draw

horizontal lines through the graph. If any horizontal line intersects the graph

more than once, then the graph does not represent a one to one function.

Given a graph of a function, use the horizontal line test to determine if the

graph represents a one to one function.

1. Inspect the graph to see if any horizontal line drawn would intersect the

curve more than once.

2. If there is any such line, determine that the function is not one to

one.**Example 15**

Applying the Horizontal Line Test

Consider the functions shown. Are either of the functions one to one?

Solution:

1. The function in part 1 is not one to one. The horizontal line shown

intersects the graph of the function at two points.

2. The function in part 2 is one to one. Any horizontal line will intersect a

diagonal line at most once.**Identifying Basic Toolkit Functions**

In this section, we will be exploring functions-the shapes of their graphs,

their unique characteristics, their algebraic formulas, and how to

solve problems with them. When learning to read, we start with the alphabet.

When learning to do arithmetic, we start with numbers. When working with

functions, it is similarly helpful to have a base set of building block

elements. We call these our toolkit functions, which form a set of basic named

functions for which we known the graph, formula, and special properties. Some

of these functions are programmed to individual buttons on many

calculators. For these definitions, we will use the x as the input variable and

y=f(x) as the output variable.

We will see these toolkit functions, combinations of toolkit functions, their

graphs, and their transformations frequently throughout this book. It

will be very helpful if we can recognize these toolkit functions and their

features quickly by name, formula, graph, and basic table properties. The

graphs and sample table values are included with each function shown.

**Domain and Range**

**Finding The Domain Of A Function**

In determining domains and ranges, we need to consider what is physically

possible in real world examples. We also need to consider what is

mathematically permitted. For example, we cannot include any input value

that leads us to take an even root of a negative number if the domain and

range consist of real numbers. Or in a function expressed as a formula,

we cannot include any input value in the domain that would lead us to divide

by 0.

We can visualize the domain as a holding area that contains raw materials

for a function machine and the range as another holding area for the

machine's products.

We can write the domain and range in interval notation, which uses

values within brackets to describe a set of numbers. In interval notation, we

use a square bracket when the set includes the endpoint and a

parenthesis to indicate the endpoint is either not included or the interval

is unbounded. For example, if a person has $100 to spend, they would need to

express the interval that is more than - and less than or equal to 100 and write

(0,100].

Let us turn our attention to finding the domain of a function whose equation

is provided. Oftentimes, finding the domain of such functions involves

remembering three different forms. First, if the function has no

denominator or an odd root, consider whether the domain could be all real

numbers. Second, if there is a denominator in the function's equation,

exclude values in the domain that force the denominator to be zero.

Third, if there is an even root, consider excluding values that would make

the radicand negative.

The smallest term from the interval is written first.

The largest term in the interval is written second, following a comma.

Parentheses are used to signify that an endpoint is not included, called

exclusive.

Brackets are used to indicate that an endpoint is included, called

inclusive.**Example 1**

Find the domain of the following function:

{(2,10), (3,10), (4,20), (5,30),(6,40)}

Solution:

First, identify the input values. The input value is the first coordinate in

an ordered pair. There are no restrictions, as the ordered pairs are simply

listed. The domain is the set of the first coordinates of the ordered pairs.

{2,3,4,5,6}**Example 2**

Find the domain of the function \(f(x)=x^2-1\)

Solution:

The input value, shown by the variable x in the equation, is squared and

then the result is lowered by 1. Any real number may be squared and then be

lowered by 1, so there are no restrictions on the domain of this function.

The domain is the set of real numbers.

In interval form, the domain of f is: (-inf,inf)**Example 3**

Find the domain of the function:

\(f(x)=\frac{x+1}{2-x}\)

Solution:

When there is a denominator, we want to include only values that do not force

the denominator to be zero. So, we will set the denominator equal to 0

and solve for x.

\(2-x=0\)

\(-x=-2\)

\(x=2\)

Now, we will exclude 2 from the domain. The answers are all real numbers

where x<2 or x>2. We can use a symbol known as the union to combine the two

sets. In interval notation, we write the solution:

\((-\infty,2) \cup (2,\infty)\)**Example 4**

Find the domain of the function \(f(x)=\sqrt{7-x}\).

Solution:

When there is an even root in the formula, we exclude any real numbers that

result in a negative number in the radicand.

Set the radicand greater than or equal to zero and solve for x

\(7-x \geq 0\)

\(-x \geq -7\)

\(x \leq 7\)

Now, we will exclude any number greater than 7 from the domain. The answers

are all real numbers less than or equal to 7.

\((-\infty,7)\)**Can there be functions in which the domain and range do not intersect at all?**

Yes, for example, the function \(f(x)=-\frac{1}{\sqrt{x}}\) has

the set of all positive real numbers as its domain but the set of all negative

real numbers as its range. As a more extreme example, a function's inputs and

outputs can be completely different categories, in such cases the domain and

range have no elements in common.**Using Notations to Specify Domain and Range**

In the previous examples, we used inequalities and lists to describe the

domain of functions. We can also use inequalities, or other statements that

might define sets of values or data, to describe the behavior of the variable

in set-builder notation. For example, \({x|10 \leq x < 30}\) describes the

behavior of x in set-builder notation. The braces "{}" are read as "the set of",

and the vertical bar "|" is read as "such that" so we would read it as "the set

of x-values such that 10 is less than or equal to x, and x is less than 30."

Here are some examples of different notation:

\(5<h\leq10\) : \({h|5<h\leq10}\) : \(5,10]\)

\(5\leq h<10\) : \({h|5\leq h<10}\) : \([5,10)\)

\(5<h<10\) : \({h|5<h<10}\) : \((5,10)\)

\(h<10\) : \({h|h<10}\) : \((-\infty,10)\)

all numbers : \({R}\) : \((-\infty,\infty)\)

To combine two intervals using inequality notation or set-builder notation, we

use the word "or". As we saw in earlier examples, we use the union symbol,

\(\cup\), to combine two unconnected intervals. For example, the union of

sets {2,3,5} and {4,6} is the set {2,3,4,5,6}. It is the set of all elements

that belong to one or the other(or both) of the original two sets. For

sets with a finite number of elements like these, the elements do not have to

be listed in ascending order of numerical value. If the original two sets

have some elements in common, those elements should be listed only once in the

union set. For sets of real numbers on intervals, another example of a

union is: \({x| |x| \geq 3} = (-\infty,-3) \cup [3,\infty)\)**Set-Builder Notation**

This is a method of specifying a set of elements that satisfy a certain

condition. It takes the form \({x| ... x}\) which reads as, "the set of all

x such that the statement about x is true."

\({x|4<x\leq12}\)**Interval Notation**

This is a way of describing sets that include all real numbers between a lower

limit that may or may not be included and an upper limit that may or may

not be included. The endpoint values are listed between brackets or

parentheses. A square bracket indicates inclusion in the set, and a

parenthesis indicates exclusion from the set.

\((4,12]\)**Given a line graph, describe the set of values using interval notation.**

1. Identify the intervals to be included in the set by determining where the

heavy line overlays the real line.

2. At the left end of each interval, use [ with each end value to be included

in the set (solid dot) or a ( for each excluded end value (open dot).

3. At the right end of each interval, use ] with each end value to be included

in the set (filled dot) or ) for each excluded end value (open dot).

4. Use the union symbol \(\cup\) to combine all intervals into one set.**Example 5**

Describe the intervals of values shown using inequality notation,

set-builder notation, and interval notation.

Solution:

To describe the values, x, included in the intervals shown, we would say "x is

a real number greater than or equal to 1 and less than or equal to 3, or a real

number greater than 5."

Inequality: \(1 \leq x \leq 3 \text{ or} x > 5\)

Set-Builder: \({x| 1 \leq x \leq 3 \text{ or} x > 5}\)

interval: \([1,3] \cup (5,\infty)\)

Remember that, when writing or reading interval notation, using a square

bracket means the boundary is included in the set. Using a parenthesis means

the boundary is not included in the set.**Finding Domain and Range From Graphs**

Another way to identify the domain and range of functions is by using graphs.

Because the domain refers to the set of possible input values, the domain of a

graph consists of all input values shown on the x-axis. The range is the

set of all possible values, which are shown on the y-axis. Keep in mind that

if the graph continues beyond the portion of the graph we can see, the domain

and range may be greater than the visible values. We can observe that the

graph extends horizontally from -5 to the right without bound, so the

domain is \([-5,\infty)\). The vertical extent of the graph is all range values

5 and below, so the range is \((-\infty, 5]. Note that the domain and range are

always written from smaller to larger values, or from left to right for domain,

and from the bottom of the graph to the top of the graph for range.**Example 6**

Find the domain and range of the function f whose graph is shown.

Solution:

We can observe that the horizontal extent of the graph is -3 to 1, so the domain

of f is (-3,1].

The vertical extent of the graph is 0 to -4, so the range is [-4,0].**Example 7**

Find the domain and range of the function f whose graph is shown.

Solution:

The input quantity along the horizontal axis is years, which we represent with

the variable T for time. The output quantity is thousands of barrels a

per day, which we represent with the variable b for barrels. The graph may

continue to the left and right beyond what is viewed, but based on the portion

of the graph that is visible, we can determine the domain as \(1973 \leq t

\leq 2008\) and the range as approximately \(180 \leq b \leq 2010\).

In interval notation, the domain is [1973,2008] and the range is about

[180,2010]. for the domain and the range, we approximate the smallest

and largest values since they do not fall exactly on the grid lines. **Can a function's domain and range be the same?**

Yes. For example, the domain and range of the cube root function are both

the set of all real numbers. **Finding Domains and Ranges of the Toolkit Functions**

We will now return to our set of toolkit functions to determine the domain and

range of each.**Constant Function **

\(f(x)=c\)

D=\((-\infty,\infty)\) : R=\([c,c]\)

For the constant function f(x)=c, the domain consists of all real numbers.

There are no restrictions on the input. The only output value is the

constant c, so the range is the set {c} that contains this single element. In

interval notation, this is written as [c,c], the interval that both begins

and ends with c.**Identity Function**

\(f(x)=x\)

D=\((-\infty,\infty)\) : R=\((-\infty,\infty)\)

For the identity function f(x)=x, there is no restriction on x. Both the

domain and and range are the set of all real numbers.**Absolute Function**

\(f(x)=|x|\)

D=\((-\infty,\infty)\) : R=\([0,\infty)\)

For the absolute value function, \(f(x)=|x|\), there is no restriction on x.

However, because absolute value is defined as a distance from 0, the

output can only be greater than or equal to 0.**Quadratic Function**

\(f(x)=x^2\)

D=\((-\infty,\infty)\) : R=\([0,\infty)\)

For the quadratic function \(f(x)=x^2\), the domain is all real numbers since

the horizontal extent of the graph is the whole real number line. Because the

graph does not include any negative value for the range, the range is only

nonnegative real numbers. **Cubic Function**

\(f(x)=x^3\)

D=\((-\infty,\infty)\) : R=\((-\infty,\infty)\)

For the cubic function \(f(x)=x^3\), the domain is all real numbers because the

horizontal extent of the graph is the whole real number line. The same

applies to the vertical extent of the graph, so the domain and range includes

all real numbers.**Reciprocal Function**

\(f(x)=\frac{1}{x}\)

D=\((-\infty,0) \cup (0,\infty)\) : R=\((-\infty,0) \cup (0,\infty)\)

We cannot divide by 0, so we must exclude 0 from the domain. further, 1

divided by any value can never be 0, so the range also will not include 0.**Reciprocal Squared Function**

\(f(x)=\frac{1}{x^2}\)

D=\((-\infty,0) \cup (0,\infty)\) : R=\((0,\infty)\)

We cannot divide by 0, so we must exclude 0 from the domain. There is also no

x that can give an output of 0, so 0 is excluded from the range as well. Note

that the output of this function is always positive due to the square in the

denominator, so the range includes only positive numbers. **Square Root Function**

\(f(x)=\sqrt{x}\)

D=\([0,\infty)\) : R=\([0,\infty)\)

We cannot take the square root of a negative real number, so the domain

must be 0 or greater. The range also excludes negative numbers because the

square root of a positive number x is defined to be positive, even though the

square of the negative number \(-\sqrt{x}\) also gives us x.**Cube Root Function**

\(f(x)=\sqrt[3]{x}\)

D=\((-\infty,\infty)\) : R=\((-\infty,\infty)\)

The domain and range include all real numbers. Note that there is no problem

taking a cube root, or any odd-integer root, of a negative number, and the

resulting output is negative(it is an odd function).**Given the formula for a function, determine the domain and range**

1. Exclude from the domain any input values that result in division by 0.

2. Exclude from the domain any input values that have nonreal or undefined

number outputs.

3. Use the valid input values to determine the range of the output values.

4. Look at the function graph and table values to confirm the actual

function behavior.**Example 8**

Find the domain and range of \(f(x)=2x^3-x\)

Solution:

There are no restrictions on the domain, as any real number may be cubed and

then subtracted from the result. The domain is \((-\infty,\infty)\) and the

range is also \((-\infty,\infty)\).**Example 9**

Find the domain and range of \(f(x)=\frac{2}{x+1}\).

Solution:

We cannot evaluate the function at -1 because division by zero is undefined. The

domain is \((-\infty,-1)\cup(-1,\infty)\). Because the function is never 0, we

exclude 0 from the range. The range is \((-\infty,0)\cup(0,\infty)\).**Example 10**

Find the domain and range of \(f(x)=2\sqrt{x+4}\).

We cannot take the square root of a negative number, so the value inside the

radical must be nonnegative.

\(x+4 \geq 0 \text{ when } x \geq -4\)

The domain of \(f(x) \text{ is } [-4,\infty)\).

We then find the range. We know that f(-4)=0, and the function value increases

as x increases without any upper limit. We conclude that the range of f is

\([0,\infty)\). **Graphing Piecewise-Defined Functions**

Sometimes, we come across a function that requires more than one formula in

order to obtain the given output. For example, in the toolkit functions, we

introduced the absolute value function \(f(x)=|x|\). With a domain of all real

numbers and a range of values greater than or equal to 0, absolute value can be

defined as the magnitude, or modulus, of a real number value regardless of sign.

It is the distance from 0 on the number line. All of these definitions require

the output to be greater than or equal to 0.

If we input 0, or a positive value, the output is the same as the input.

\(f(x)=x \text{ if } x \geq 0\)

If we input a negative value, the output is the opposite of the input.

\(f(x)=-x \text{ if } x < 0\)

Because this requires two different processes or pieces, the absolute

value function is an example of a piecewise function. A piecewise function is a

function in which more than one formula is used to define the output over

different pieces of the domain.

We use piecewise functions to describe situations in which a rule or

relationship changes as the input value crosses certain boundaries. For

example, we often encounter situations in business for which the cost per piece

of a certain item is discounted once the number ordered exceeds a certain

value. Tax brackets are another real-world example of piecewise functions. For

example, consider a simple tax system in which incomes up to $10,000 are taxed

at 10%, and any additional income is taxed at 20%. The tax on a total income S

would be \(0.1S \text{ if } S \leq 10,000 \text{ and } 1000+0.2(S-10000) \text{

if } S > 1000\).**Piecewise Function**

A piecewise function is a function in which more than one formula is used to

define the output. Each formula has its own domain, and the domain of the

function is the union of all these smaller domains. **Given a piecewise function, write the formula and identify the domain for****each interval.**

1. Identify the intervals for which different rules apply.

2. Determine formulas that describe how to calculate an output from an input

in each interval.

3. Use braces and if-statements to write the function.**Example 11**

Writing a Piecewise Function

A museum charges $5 per person for a guided tour with a group of 1-9 people or

a fixed $50 fee for a group of 10 or more people. Write a function relating

the number of people, n, to the cost, c.

Solution:

Two different formulas will be needed. For n-values under 10, c=5n. For

values of n that are 10 or greater, c=50.

\[c(n)= \begin{cases} 5n \text{ if } 0 < n < 10 & \quad \\ 50 \text{ if } n

\geq 10 & \quad \end{cases}\]**Example 12**

A cell phone company uses the function below to determine the cost, c, in

dollars for g gigabytes of data transfer.

\[c(g)= \begin{cases} 25 \text{ if } 0 < g < 2 & \quad \\ 25 + 10(g-2) \text{ if

} g \geq 2 & \quad \end{cases}\]

Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes

of data.

Solution:

To find the cost of using 1.5 gigabytes of data, c(1.5), we first look to see

which part of the domain our input falls in. Because 1.5 is less than 2, we use

the first formula.

\(c(1.5)=$25\)

To find the cost of using 4 gigabytes of data, c(4), we see that our input of

4 is greater than 2, so we use the second formula.

\(c(4)=25+10(4-2)=$45\)**Given a piecewise function, sketch a graph**

1. Indicate on the x-axis the boundaries defined by the intervals on each

piece of the domain.

2. For each piece of the domain, graph on that interval using the

corresponding equation pertaining to that piece. Do not graph two functions

over one interval because it would violate the criteria of a function.**Example 13**

Sketch a graph of the function:

\[f(x)=\begin{cases} x^2 \text{ if } x \leq 1 & \quad \\ 3 \text{ if } 1 < x

\leq 2 & \quad \\ x \text{ if } x > 2 & \quad \end{cases}\]

Solution:

Each of the component functions is from our library of toolkit functions, so we

know their shapes. We can imagine graphing each function and then

limiting the graph to the indicated domain. At the endpoints of the domain,

we draw open circles to indicate where the endpoint is not included because of a

less-than or greater-than inequality. We draw a closed circle where the endpoint

is included because of a less-than-or-equal-to or greater-than-or-equal-to

inequality.**Can more than one formula from a piecewise function be applied to a value in the****domain?**

No. Each value corresponds to one equation in a piecewise formula.

**Rates of Change**

**Finding the Average Rate of Change of a Function**

The price change per year is a rate of change because it describes how an

output quantity changes relative to the change in the input quantity. We can

see that the price of gasoline in the table did not change by the same amount

each year, so the rate of change was not constant. If we use only the beginning

and ending data, we would be finding the average rate of change over the

specified period of time. To find the average rate of change, we divide the

change in the output value by the change in the input value.

Average rate of change=change in output/change in input

\(=\frac{\Delta y}{\Delta x}\)

\(\frac{y_2 - y_1}{x_2 - x_1}\)

\(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\)

The Greek letter \(\Delta\) (delta) signifies the change in a quantity. We

read the ratio as "delta-y over delta-x" or "the change in y divided by

the change in x". Occasionally we write \(\Delta f\) instead of \(\Delta y\),

which still represents the change in the function's output value resulting

from a change to its input value. It does not mean we are changing the function

into some other function.

In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7

years, the average rate of change was:

\(\frac{\Delta y}{\Delta x} = \frac{$1.37}{7years} = 0.196 \text{ dollars per

year }\)

On average, the price of gas increased by about 19.6 cents per years.**Rate Of Change**

A rate of change describes how an output quantity changes relative to the

change in the input quantity. The units on a rate of change are "output units

per input units".

The average rate of change between two input values is the total change of the

function values (output values) divided by the change in the input values.

\(\frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\)

Given the value of a function at different points, calculate the average

rate of change of a function for the interval between two values x1 and x2.

1. Calculate the difference of \(y_2-y_1=\Delta y\)

2. Calculate the difference of \(x_2-x_1=\Delta x\)

3. Find the ratio \(\frac{\Delta y}{\Delta x}\)**Example 1**

Find the average rate of change of the price of gasoline between 2007 and

2009.

Solution:

In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The

average rate of change is:

\(\frac{\Delta y}{\Delta x} = \frac{2.41-2.84}{2009-2007} = \frac{-0.43}{2} =

-$0.22\) per year

Note that a decrease is expressed by a negative change or negative increase. A

rate of change is negative when the output decreases as the input increases or

when the output increases as the input decreases.**Example 2**

Computing average rate of change from a graph

Given the function g(t), find the average rate of change on the interval

[-1,2].

Solution:

At t=-1, the graph shows g(-1) =4. At t=2, the graph shows g(2)=1.

The horizontal change \(\Delta t= 3\) is shown by the red arrow, and the

vertical change \(\Delta g(t)=-3\) is shown by the blue arrow. The output

changes by -3 while the input changes by 3, giving an average rate of change

of:

\(\frac{1-4}{2-(-1)} = \frac{-3}{3} = -1 \)

Note that the order we choose is very important. If, for example, we use

\(\frac{y_2-y_1}{x_1-x_2}\), we will not get the correct answer. Decide

which point will be 1 and which point will be 2, and keep the coordinates

fixed as \((x_1,y_1)\) and \(x_2,y_2)\).**Example 3**

Computing Average Rate of Change from a Table

After picking up a friend who lives 10 miles away, anna records her distance

from home over time. The values are shown in the table. Find her average

speed over the first 6 hours.

t(hours) =0 1 2 3 4 5 6 7

D(t)(miles) =10 55 90 153 214 240 292 300

Solution:

Here, the average speed is the average rate of change. She traveled 282 miles in

6 hours, for an average speed of:

\(\frac{292-10}{6-0} = \frac{282}{6} = 47\)

Because the speed is not constant, the average speed depends on the interval

chosen. For the interval[2,3], the average speed is 63 miles per hour.**Example 4**

Computing Average Rate of Change for a Function Expressed as a Formula

Compute the average rate of change of \(f(x)=x^2-\frac{1}{x}\) on the

interval [2,4].

Solution:

We can start by computing the function values at each endpoint of the interval.

\(f(2)=2^2-1/2 = 4-1/2 = 7/2\)

\(f(4)=4^2-1/4 = 16-1/4 = 63/4\)

Now we compute the average rate of change:

\(\frac{f(4)-f(2)}{4-2} = 49/8\)**Example 5**

Finding the Average Rate of Change of a Force

The electrostatic force F, measured in newtons, between two charged

particles can be related to the distance between the particles d, in

centimeters, by the formula \(F(d)=\frac{2}{d^2}\). Find the average rate of

change of force if the distance between the particles is increased from 2 cm to

6 cm.

Solution:

We are computing the average rate of change of \(F(d)=\frac{2}{d^2}\) on

the interval [2,6].

Average rate of change:

\(\frac{F(6)-F(2)}{6-2} = \frac{(2/36)-(2/4)}{6-2} = \frac{-(16/36)}{4} =

-(1/9)\)

The average rate of change is -(1/9) newton per centimeter.**Example 6**

Finding an Average Rate of Change as an Expression

Find the average rate of change of \(g(t)=t^2+3t+1\) on the interval [0,a].

The answer will be an expression involving a.

Solution:

We use the average rate of change formula.

\(\frac{g(a)-g(0)}{a-0}\)

\(\frac{(a^2+3a+1)-(0^2+3(0)+1)}{a-0}\)

\(\frac{a^2+3a+1-1}{a} = \frac{a(a+3)}{a}=a+3\)

This result tells us the average rate of change in terms of "a" between t=0

and any other point t=a. For example, on the interval [0,5], the average rate

of change would be 5+3=8.**Using a Graph to Determine Where a Function is Increasing, Decreasing,****or Constant**

As part of exploring how functions change, we can identify intervals over

which the function is changing in specific ways. We say that a function is

increasing on an interval if the function values increase as the input

increases within that interval. Similarly, a function is decreasing on an

interval if the function values decrease as the input values increase over that

interval. The average rate of change of an increasing function is positive, and

the average rate of change of a decreasing function is negative.

While some functions are increasing or decreasing over their entire domain, many

others are not. A value of the input where a function changes from increasing

to decreasing (as we go from left to right, that is, as the input variable

increases) is the location of a local maximum. The function value at that

point is the local maximum. If a function has more than one, we say it has

local maxima. Similarly, a value of the input where a function changes from

decreasing to increasing as the input variable increases is the location

of a local minimum. The function value at that point is the local minimum. The

plural form is local minima. Together, local maxima and local minima are

called local extrema, or local extreme values, of the function. The singular

form is extrenum. Often, the term local is replaced by the term relative.

Clearly, a function is neither increasing nor decreasing on an interval

where it is constant. A function is also neither increasing nor decreasing

at extrema. Note that we have to speak of local extrema, because any

given local extrenum as defined here is not necessarily the highest

maximum or lowest minimum in the function's entire domain.

For the function whose graph is shown, the local maximum is 16, and it

occurs at x=-2. The local minimum is -16 and it occurs at x=2.

To locate the local maxima and minima from a graph, we need to observe the

graph to determine where the graph attains its highest and lowest points,

respectively, within an open interval. Like the summit of a roller coaster, the

graph of a function is higher at a local maximum than at nearby points on both

sides. The graph will also be lower at a local minimum than at neighboring

points. These observations lead us to a formal definition of local

extrema.**Local Minima and Local Maxima**

A function f is an increasing function on an open interval if f(b)>f(a) for

every two input values a and b in the interval where b>a.

A function f is a descending function on an open interval if f(b)<f(a) for

every two input values a and b in the interval where b>a.

A function f has a local maximum at a point b in an open interval (a,c) if

\(f(b) \geq f(x)\) for every point x (x does not equal b) in the interval. F has

a local minimum at a point b in (a,c) if \(f(b) \leq f(x)\) for every point x

(x does not equal b) in the interval.**Example 7**

Finding increasing and Decreasing intervals on a Graph

Given the function p(t), identify the intervals on which the function

appears to be increasing.

Solution:

We see that the function is not constant on any interval. The function is

increasing where it slants upward as we move to the right and decreasing

where it slants downward as we move to the right. The function appears to be

increasing from t=1 to t=3 and from t=4 on.

In interval notation, we would say the function appears to be increasing

on the interval (1,3) and the interval (4,\infty).

Notice in this example that we used open intervals (intervals that do not

include the endpoints), because the function is neither increasing nor

decreasing at t=1, t=3, and t=4. These points are the local extrema(two minima

and a maximum).**Example 8**

Finding Local Extrema from a Graph

Graph the function \(f(x)=\frac{2}{x} + \frac{x}{3}\). Then use the graph to

estimate the local extrema of the function and to determine the intervals

on which the function is increasing.

Using technology, it appears there is a low point, or local minimum, between

x=2 and x=3, and a mirror-image high point, or local maximum, somewhere

between x=-3 and x=-2.

Most graphing calculators and graphing utilities can estimate the location of

maxima and minima. **Example 9**

Finding Local Maxima and Minima from a Graph

For the function f whose graph is shown, find all local maxima nd minima.

Solution:

Observe the graph of f. The graph attains a local maximum at x=1 because it

is the highest point in an open interval around x=1. The local maximum is the

y-coordinate at x=1, which is 2.

The graph attains a local minimum at x=-1 because it is the lowest point in an

open interval around x=-1. The local minimum is the y-coordinate x=-1, which is

-2.**Analyzing the Toolkit Functions for Increasing or Decreasing Intervals**

Constant Function

\(f(x)=c\)

Neither increasing nor decreasing

Identity Function

\(f(x)=x\)

Increasing

Quadratic Function

\(f(x)=x^2\)

Increasing on (0,inf)

Decreasing on (-inf,0)

Minimum at x=0

Cubic Function

\(f(x)=x^3\)

Increasing

Reciprocal Function

\(f(x)=\frac{1}{x}\)

Decreasing (-inf,0) union (0,inf)

Reciprocal Squared Function

\(f(x)=\frac{1}{x^2}\)

Increasing on (-inf,0)

Decreasing on (0,inf)

Cube Root Function

\(f(x)=\sqrt[3]{x}\)

Increasing

Square Root Function

\(f(x)=\sqrt{x}\)

Increasing on (0,inf)

Absolute Value Function

\(f(x)=|x|\)

Increasing on (0,inf)

Decreasing on (-inf,0)**Use a Graph to Locate the Absolute Maximum and Absolute Minimum**

There is a difference between locating the highest and lowest points on a graph

in a region around an open interval(locally) and locating the highest and

lowest points on the graph for the entire domain. The y-coordinates (output) at

the highest and lowest points are called the absolute maximum and absolute

minimum, respectively.

To locate absolute maxima and minima from a graph, we need to observe the graph

to determine where the graph attains its highest and lowest points on the

domain of the function. Not every function has an absolute maximum or

minimum value. The function \(f(x)=x^3\) is one such function. **Absolute Maxima and Minima**

The absolute maximum of f at x=c is f(c) where \(f(c) \geq f(x)\) for all x in

the domain of f.

The absolute minimum of f at x=d is f(d) where \(f(d) \leq f(x)\) for all x in

the domain of f.**Example 10**

Finding Absolute Maxima and Minima from a Graph

For the function f, find all absolute maxima and minima.

Solution:

Observe the graph of f. The graph attains an absolute maximum in two locations,

x=-2 and x=2, because at these locations, the graph attains its highest point

on the domain of the function. The absolute maximum is the y-coordinate at x=-2

and x=2, which is 16.

The graph attains an absolute minimum at x=3, because it is the lowest point on

the domain of the function's graph. The absolute minimum is the

y-coordinate at x=3, which is -10.

**Composition of Functions**

**Combining Functions Using Algebraic Operations**

Function composition is only one way to combine existing functions.

Another way is to carry out the usual algebraic operations on functions,

such as addition, subtraction, multiplication, and division. We do all this

by performing the operations with the function outputs, defining the

result as the output of our new function.

Suppose we need to add two columns of numbers that represent a husband and

wife's separate annual incomes over a period of years, with the result being

their total household income. We want to do this for every year, adding only

that year's incomes and then collecting all the data in a new column. If w(y)

is the wife's income and h(y) is the husband's income in year y, and we want T

to represent the total income, then we can define a new function.

\(T(y)=h(y)+w(y)\)

If this holds true for every year, then we can focus on the relation between

the functions without reference to a year and write:

\(T=h+w\)

Just as for this sum of two functions, we can define difference, product, and

ratio functions for any pair of functions that have the same kind of inputs

(not necessarily numbers) and also the same kinds of outputs (which do have to

be numbers so that the usual operations of algebra can apply to them, and which

also must have the same units or no units when we add and subtract). In this

way, we can think of adding, subtracting, multiplying, and dividing functions.

For two functions f(x) and g(x) with real number outputs, we define new

functions f+g, f-g, fg, and f/g by the relations:

\((f+g)(x)=f(x)+g(x)\)

\((f-g)(x)=f(x)-g(x)\)

\((fg)(x)=f(x)g(x)\)

\((f/g)(x)=f(x)/g(x)\)**Example 1**

Performing Algebraic Operations on Functions

Find and simplify the functions \((g-f)(x)\) and \((g/f)(x)\), given

\(f(x)=x-1\) and

\(g(x)=x^2-1\). Are they the same function?

Solution:

begin by writing the general form, and then substitute the given functions.

\((g-f)(x) = g(x)-f(x)\)

\((g-f)(x) = x^2-1-(x-1)\)

\((g-f)(x) = x^2-x\)

\((g-f)(x) = x(x-1)\)

\((g/f)(x) = \frac{g(x)}{f(x)}\)

\((g/f)(x) = \frac{x^2-1}{x-1}\)

\((g/f)(x) = \frac{(x+1)(x-1)}{x-1}\)

\((g/f)(x) = x+1\)

No, the functions are not the same.

Note, for (g/f)(x), the condition \(x \not = 1\) is necessary because when

x=1, the denominator is equal to 0, which makes the function undefined.**Create a Function by Composition of Functions**

Performing algebraic operations on functions combines them into a new function,

but we can also create functions by composing functions. When we wanted

to compute a heating cost from a day of the year, we created a new function

that takes a day as input and yields a cost as output. The process of combining

functions so that the output of one function becomes the input of another is

known as the composition of functions. The resulting function is known as a

composition function. We represent this combination by the following notation:

\(f \circ g)(x)=f(g(x))\)

We read the left-hand side as "f composed with g at x", and the right=hand side

as "f of g of x". The two sides of the equation have the same mathematical

meaning and are equal. The open circle is called the composition operator. We

use this operator mainly when we wish to emphasize the relationship between the

functions themselves without referring to any particular input value.

Composition is a binary operation that takes two functions and forms a new

function, much as addition or multiplication takes two numbers and gives a new

number. However, it is important not to confuse function composition with

multiplication because, as we learned above, in most cases

\(f(g(x))\not=f(x)g(x)\).

It is also important to understand the order of operations in evaluating a

composite function. We follow the usual convention with parentheses by starting

with the innermost parentheses first, and then working to the outside. In the

equation above, the function g takes the input x first and yields an output

g(x). Then the function f takes g(x) as an input and yields an output f(g(x)).

g(x), the output of g is the input of f

In general, \(f \circ g\) and \(g \circ f\) are different functions. in other

words, in many cases, \(f(g(x)) \not= g(f(x))\) for all x. We will also see that

sometimes two functions can be composed only in one specific order.

For example, if \(f(x)=x^2\) and \(g(x)=x+2\), then:

\(f(g(x))=f(x+2) = (x+2)^2 = x^2+4x+4\)

but:

\(g(f(x))=g(x^2) = x^2+2\)

These expressions are not equal for all values of x, so the two functions are

not equal. It is irrelevant that the expressions happen to be equal for the

single input value \(x=1/2\).

Note that the range of the inside function (first function to be evaluated)

needs to be within the domain of the outside function. Less formally, the

composition has to make sense in terms of inputs and outputs.**Composition of Functions**

When the output of one function is used as the input of another function, we

call the entire operation a composition of functions. For any input x and

functions f and g, this action defines a composite function, which we write as

\(f \circ g\) such that \((f \circ g)(x)=f(g(x))\).

The domain of the composite function \(f \circ g\) is all x such that x is in

the domain of g and g(x) is in the domain of f. It is important to realize that

the product of functions fg is not the same as the function composition f(g(x)),

because , in general, \(f(x)g(x)\not=f(g(x))\).**Example 2**

Determining Whether Compositions of Functions is Commutative

Using the functions provided, find f(g(x)) and g(f(x)). Determine whether the

composition of the functions is commutative.

\(f(x)=2x+1\) and \(g(x)=3-x\)

Solution:

Let's begin by substituting g(x) into f(x).

\(f(g(x))=2(3-x)+1 = 6-2x+1 = 7-2x\)

Now, we can substitute f(x) into g(x).

\(g(f(x))=3-(2x+1) = 3-2x-1 = -2x+2\)

We find that \(g(f(x)) \not= f(g(x))\), so the operation of function composition

is not commutative.**Example 3**

Interpreting Composite Functions

The function c(s) gives the number of calories burned completing s sit-ups, and

s(t) gives the number of sit-ups a person can complete in t minutes. Interpret

c(s(3)).

Solution:

The inside expression in the composition is s(3). Because the input to the

s-function is time, t=3 represents 3 minutes, and s(3) is the number of sit-ups

completed in 3 minutes.

Using s(3) as the input to the function c(s) gives us the number of calories

burned during the number of sit-ups that can be completed in 3 minutes, or

simply the number of calories burned in 3 minutes by doing sit-ups. **Example 4**

Investigating the Order of Function Composition

Suppose f(x) gives miles that can be driven in x hours and g(y) gives the

gallons of gas used in driving y miles. Which of the expressions is meaningful,

f(g(y)) or g(f(x))?

Solution:

The function \(y=f(x)\) is a function whose output is the number of miles driven

corresponding to the number of hours driven.

number of miles=f(number of hours)

The function g(y) is a function whose output is the number of gallons used

corresponding to the number of miles driven. This means:

number of gallons = g(number of miles)

The expression g(y) takes miles as the input and a number of gallons as the

output. The function f(x) requires a number of hours as the input. Trying to

input a number of gallons does not make sense. The expression f(g(y)) is

meaningless.

The expression f(x) takes hours as the input and a number of miles driven as the

output. The function g(y) requires a number of miles as the input. Using

f(x)(miles driven) as an input value for g(y), where gallons of gas depends on

miles driven, does make sense. The expression g(f(x)) makes sense, and will

yield the number of gallons of gas used, g, driving a certain number of miles,

f(x), in x hours.**Are there any situations where f(g(y)) and g(f(x)) would both be meaningful or****useful expressions?**

Yes. For many pure mathematical functions, both compositions make sense, even

though they usually produce different new functions. In real-world problems,

functions whose inputs and outputs have the same units also may give

compositions that are meaningful in either order.**Evaluating Composite Functions**

Once we compose a new function from two existing functions, we need to be able

to evaluate it for any input in its domain. We will do this with specific

numerical inputs for functions expressed as tables, graphs, and formulas and

with variables as inputs to functions expressed as formulas. In each case, we

evaluate the inner function using the starting input and then use the inner

function's output as the input for the outer function.

When working with functions given as tables, we read input and output values

from the table entries and always work from the inside to the outside. We

evaluate the inside function first and then use the output as of the inside

function as the input to the outside function.**Example 5**

Using a Table to Evaluate a Composite Function

Evaluate f(g(3)) and g(f(3)).

x f(x) g(x)

1 6 3

2 8 5

3 3 2

4 1 7

Solution:

To evaluate f(g(3)), we start from the inside with the input value 3. We then

evaluate the inside expression g(3) using the table that defines the function

g:g(3)=2. We can then use that result as the input to the function f, sp g(3) is

replaced by 2 and we get f(2). Then using the table that defines the function f,

we find that f(2)=8.

\(g(3)=2\)

\(f(g(3))=f(2)=8\)

To evaluate g(f(3)), we first evaluate the inside expression f(3) using the

first table: f(3)=3. Then, using the table for g, we can evaluate:

\(g(f(3))=g(3)-2\)**Evaluating Composite Functions Using Graphs**

When we are given individual functions as graphs, the procedure for evaluating

composite functions is similar to the process we use for evaluating tables. We

read the input and output values, but this time, from the x and y axes of the

graphs.

Given a composite function and graphs of its individual functions, evaluate it

using the information provided by the graphs.

1. Locate the given input to the inner function on the x-axis of its graph

1. Read off the input of the inner function from the y-axis of its graph

3. Locate the inner function output on the x-axis of the graph of the outer

function.

4. Read the output of the outer function from the y-axis of its graph. This is

the output of the composite function.**Example 6**

Using a Graph to Evaluate a Composite Function

Evaluate f(g(1))

Solution:

We evaluate g(1) using the graph of g(x), finding the input of 1 on the x-axis

and finding the output value of the graph at that input. Here, g(1)=3. We

use this value as the input to the function f.

\(f(g(1))=f(3)\)

We can then evaluate the composite function by looking to the graph of f(x),

finding the input of 3 on the x-axis and reading the output value of the graph

at this input. Here, \(f(3)=6, \text{ so } f(g(1))=6\).**Evaluating Composite Functions Using Formulas**

When evaluating a composite function where we have either created or been

given formulas, the rule of working from the inside out remains the same. The

input value to the outer function will be the output of the inner function,

which may be a numerical value, a variable name, or a more complicated

expression.

While we can compose the functions for each individual input value, it is

sometimes helpful to find a single formula that will calculate the result of a

composition \(f9g(x))\). To do this, we will extend our idea of the function

evaluation. Recall that, when we evaluate a function like \(f(t)=t^2-t\), we

substitute the value inside the parentheses into the formula wherever we see

the input variable.

Given a formula for a composite function, evaluate the function.

1. Evaluate the inside out function using the input value or variable

provided.

2. use the resulting output as the input to the outside function.**Example 7**

Evaluating a Composition of Functions Expressed as Formulas with a numerical

Input

Given \(f(t)=t^2-t\) and \(h(x)=3x+2\), evaluate f(h(1)).

Solution:

Because the inside expression is h(1), we start by evaluating h(x) at 1.

\(h(1)=3(1)+2 = 5\)

Then \(f(h(1))=f(5)\), so we evaluate f9t) at an input of 5.

\(f(h(1))= f(5) = 5^2-5 = 20\)

It makes no difference what the input variables t and x were called because we

evaluated for specific numerical values.**Finding the Domain of a Composite Function**

As we discussed previously, the domain of a composite function such as \(f

\circ g\) is dependent on the domain of g and the domain of f. It is

important to know when we can apply a composite function and when we

cannot, that is, to know the domain of a function such as \(f \circ g\). Let us

assume we know the domains of the functions f and g separately. If we write

the composite function for an input x as f(g(x)), we can see right away that x

must be a member of the domain of g in order for the expression to be

meaningful, because otherwise we cannot complete the inner function

evaluation. However, we can see right away that x must be a member of the domain

of f, otherwise the second function evaluation in f(g(x)) cannot be

completed, and the expression is still undefined. Thus the domain of \(f \circ

g\) consists of only those inputs in the domain of g that produce outputs from

g belonging to the domain of f. Note that the domain of f composed with g is

the set of all x such that x is in the domain of f composed with g is the

set of all x such that x is in the domain of g and g9x) is in the domain of

f.**Domain of a Composite Function**

The domain of a composite function f(g(x)) is the set of those inputs x in the

domain of g for which g(x) is in the domain of f.

Given a function composition f(g(x)), determine its domain

1. Find the domain of g

2. Find the domain of f

3. Find those inputs x in the domain of g for which g(x) is in the domain of

f. That is, exclude those inputs x from the domain of g for which g(x) is not

in the domain of f. The resulting set is the domain of \(f \circ g\).**Example 8**

Finding the Domain of a Composite Function

Find the domain of \((f \circ g)(x)\) where \(f(x)=5/(x-1) \text{ and } g(x) =

4/(3x-2)\).

Solution:

The domain of g(x) consists of all real numbers except \(x=2/3\), since that

input value would cause us to divide by 0. Likewise, the domain of f consists

of all real numbers except 1. So we need to exclude from the domain of g(x)

that value of x for which g9x)=1.

\(\frac{4}{3x-2} = 1\)

\(4=3x-2\)

\(6=3x = 2\)

So the domain of \(f \circ g\) is the set of all real numbers except 2/3 and 2.

This means that :

\(x \not= \frac{2}{3} \text{ or } x\not=2\)

We can write this in interval notation as:

\((-\infty,\frac{2}{3}) \cup (\frac{2}{3},2) \cup (2,\infty)\)**Example 9**

Finding the Domain of a Composite Function Involving Radicals

Find the domain of :

\((f \circ g)(x) \text{ where } f(x) = \sqrt{x+2} \text{ and } g(x) =

\sqrt{3-x}\)

Solution:

Because we cannot take the square root of a negative number, the domain of g

is \((-\infty,3]\). Now we can check the domain of the composite function:

\((f \circ g)(x) = \sqrt{\sqrt{3-x}+2}\)

This expression must \(\geq 0\), since the radicand of a square root must be

positive. Since square roots are positive, \(\sqrt{3-x} \geq 0\) which gives

the domain of \((-\infty,3]\).

This example shows that knowledge of the range of functions (specifically

the inner function) can also be helpful in finding the domain of a composite

function. It also shows that the domain of \(f \circ g\) can contain values

that are not in the domain of f, though they must be in the domain of g.

Decomposing a Composite Function into its Component Functions

In some cases, it is necessary to decompose a complicated function. In

other words, we can write it as a composition of two simpler functions. There

may be more than one way to decompose a composite function, so we may

choose the decomposition that appears to be most expedient.**Example 10**

Decomposing a Function

Write \(f(x)=\sqrt{5-x^2}\) as the composition of two functions.

We are looking for two functions, g and h, so f(x)=g(h(x)). To do this,

we look for a function inside a function in the formula for f(x). As one

possibility, we might notice that the expression \(5-x^2\) is the inside of

the square root. We could then decompose the function as \(h(x)=5-x^2 \text{

and } g(x) = \sqrt{x}\)

We can check out answer by recomposing the functions.

\(g(h(x))=g(5-x^2)=\sqrt{5-x^2}\)

**Transformations of Functions**

**Graphing Functions Using Vertical and Horizontal Shifts **

Often, when given a problem, we try to model the scenario using

mathematics in the form of words, tables, graphs, and equations. One method we

can employ is to adapt the basic graphs of the toolkit functions to build new

models for a given scenario. There are systematic ways to alter functions to

construct appropriate models for the problems we are trying to solve.**Identifying Vertical Shifts**

One simple kind of transformation involves shifting the entire graph of a

function up, down, right, or left. The simplest shift is a vertical shift,

moving the graph up or down, because this transformation involves

adding a positive or negative constant to the function. In other words,

we add the same constant to the output value of the function regardless of

the input. For a function \(g(x)=f(x)+k\), the function f(x) is shifted

vertically k units. To help visualize the concept of a vertical shift, consider

that \(y=f(x)\). Therefore, \(f(x)+k\) is equivalent to \(y+k\). Every unit of

y is replaced by \(y+k\), so the y-value increases or decreases depending on

the value of k. The result is a shift upward or downward.**Vertical Shift**

Given a function, f(x), a new function g(x)=f(x)+k where k is a constant, is

a vertical shift of the function f(x). All the output values change by k units.

If k is positive, the graph will shift up. If k is negative, the graph will

shift down. **Example 1**

Adding a Constant to a Function

To regulate temperature in a green building, airflow vents near the roof open

and close throughout the day. During the summer, the facilities manager

decides to try to better regulate temperature by increasing the amount of

open vents by 20 square feet throughout the day and night. Sketch a graph of

this new function.

Solution:

We can sketch a graph of this new function by adding 20 to each of the output

values of the original function. This will have the effect of shifting the

graph vertically up. Notice that for each input value, the output value has

increased by 20, so if we call the new function S(t), we could write

\(S(t)=V(t)+20\).

This notation tells us that, for any value of t, S(t) can be found by

evaluating the function V at the same input and then adding 20 to the result.

This defines S as a transformation of the function V, in this case a vertical

shift up 20 units. Notice that, with a vertical shift, the input values stay the

same and only the output values change.**Given a tabular function, create a new row to represent a vertical****shift.**

1. Identify the output row or column

2. Determine the magnitude of the shift

3. Add the shift to the value in each output cell. Add a positive value for up

or a negative value for down.**Example 2**

Shifting a Tabular Function Vertically

Solution:

The formula g(x)=f(x)-3 tells us that we can find the output values of g by

subtracting 3 from the output values of f.

\(f(2)=1\)

\(g(x)=f(x)-3\)

\(g(2)=f(2)-3\)

\(1-3=-2\)

As with the earlier vertical shift, notice the input values stay the same and

only the output values change.**Identifying Horizontal Shifts**

We just saw that the vertical shift is a change to the output, or outside of the

function. We will now look at how changes to input, on the inside of the

function, change its graph and meaning. A shift to the input results in a

movement of the graph of the function left or right in what is known as a

horizontal shift.

For example, if \(f(x)=x^2\). then \(g(x)=(x-2)^2\) is a new function. Each

input is reduced by 2 prior to squaring the function. The result is that the

graph is shifted 2 units to the right, because we would need to increase

the prior input by 2 units to yield the same output value as given in f.

Horizontal Shift

Given a function, f, a new function g(x)=f(x-h), where h is a constant, is a

horizontal shift of the function f. If h is positive the graph will shift to

the right. If h is negative then the graph will shift to the left. **Example 3**

Adding a Constant to an Input

Returning to our building airflow example, suppose that in autumn the

facilities manager decides that the original venting plan starts too late, and

wants to begin the entire venting program 2 hours earlier. Sketch a graph

of the new function.

Solution:

We can set V(t) to be the original program and F(t) to be the revised program.

V(t)= the original venting plan

F(t)= starting 2 hours sooner

in the new graph, at each time, the airflow is the same as the original

function V, was 2 hours later. For example, in the original function V, the

airflow starts to change at 8 am, whereas for the function F, the airflow

starts to change at 6 am. The comparable function values are V(8)=F=(6).

Notice also that the vents first opened top \(220 ft^2\) at 10 am under the

original plan, while under the new plan the vents reach \(220 ft^2\) at

8am, so V(10)=F(8). In both cases, we see that,. because F(t) starts 2

hours sooner, h=-2. That means that the same output values are reached when:

\(F(t)=V(t)-(-2)) = V(t+2)\)

Note that V(t+2) has the effect of shifting the graph to the left. Horizontal

changes affect the domain of a function instead of the range and often seem

counterintuitive. The new function F(t) uses the same outputs as V(t), but

matches those outputs to inputs 2 hours earlier than those of V(t). Said

another way, we must add 2 hours to the input of V to find the corresponding

output of F: F(t)=V(t+2).**Given a tabular function, create a new row to represent a horizontal****shift.**

1. Identify the input row or column

2. Determine the magnitude of the shift

3. Add the shift to the value in each input cell.**Example 4**

Shifting a Tabular Function Horizontally

A function f(x) is given, create a table for the function g(x)=f(x-3).

Solution:

The formula \(g(x)=f(x-3)\) tells us that the output values of g are the same

as the output value of f when the input value is 3 less than the original

value. for example, we know that f(2)=1. To get the same output from the

function g, we will need an input value that is 3 larger. We input a value

that is 3 larger for g(x) because the function takes 3 away before evaluating

the function f.

\(g(5)=f(5-3)=f(2)=1\)

The result is that the function g(x) has been shifted to the right by 3. Notice

the output values for g(x) remain the same as the output values for f(x),

but the corresponding values, x, have shifted to the right by 3. Specifically,

2 shifted to 5,4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.**Example 5**

Identifying a Horizontal Shift of a Toolkit function

The graph represents a transformation of the toolkit function \(f(x)=x^2\).

Relate this new function g(x) to f(x), and then find a formula for g(x).

Solution:

Notice that the graph is identical in shape to the \(f(x)=x^2\) function,

but the x-values are shifted to the right 2 units. The vertex used to be at

(0,0), but now the vertex is at (2,0). The graph is the basic quadratic

function shifted 2 units to the right.

\(g(x)=f(x-2)\)

Notice how we must input the value x=2 to get the output value y=0. The

x-values must be 2 units larger because of the shift to the right by 2 units.

We can then use the definition of the f(x) function to write a formula for g(x)

by evaluating f(x-2).

\(F(x)=x^2\)

\(g(x)=f(x-2)\)

\(g(x)=f(x)-20=(x-2)^2\)

To determine whether the shift is +2 or -2, consider a single reference point on

the graph. For a quadratic, looking at the vertex point is convenient. In

the original function, f(0)=0. In out shifted function, g(2)=0. To

obtain the output value 0 from the function f, we need to decide whether a plus

or minus sign will work to satisfy g(2)=f(x-2)=f(0)=0. For this to work, we will

need to subtract 2 units form out input values.**Example 6**

Interpreting Horizontal versus Vertical Shifts

The function G(m) gives the number of gallons of gas required to drive m

miles. Interpret G(m)+10 and G(m+10).

Solution:

G(m)+10 can be interpreted as adding 10 to the output, gallons. This is the gas

required to drive m miles, plus another 10 gallons of gas. The graph would

indicate a vertical shift.

G(m+10) can be interpreted as adding 10 to the input, miles. So this is the

number of gallons of gas required to drive 10 miles more than m miles. The

graph would indicate a horizontal shift.**Combining Vertical and Horizontal Shifts**

Now that we have two transformations, we can combine them together. Vertical

shifts are outside changes that affect the input x axis values and shift

the function left or right. Combining the two types of shifts will cause the

graph of a function to shift up or down and right or left.

Given a function and both a vertical and a horizontal shift, sketch the graph.

1. Identify the vertical and horizontal shifts from the formula.

2. The vertical shift results from a constant added to the output. Move the

graph up for a positive constant and down for a negative constant.

3. The horizontal shift results from a constant added to the input. Move the

graph left for a positive constant and right for a negative constant.

4. Apply the shifts to the graph in either order.**Example 7**

Graphing Combined Vertical and Horizontal Shifts

Given f(x)=|x|, sketch a graph of h(x)=f(x+1)-3

Solution:

The function f is our toolkit absolute value function. We know that this graph

has a v shape, with the point at the origin. The graph of h has transformed f in

two ways: f(x+1) is a change on the inside of the function, giving a

horizontal shift left by 1, and the subtraction by 3 in f(x+1)-3 is a change

to the outside of the function, giving a vertical shift down by 3.**Example 8**

Identifying Combined Vertical and Horizontal Shifts

Write a formula for the graph shown, which is a transformation of the

toolkit square root function.

Solution:

The graph of the toolkit function starts at the origin, so this graph has

been shifted 1 to the right and up 2. In function notation, we could write

that as \(h(x)=f(x-2)+2\). Using the formula for the square root function, we

can write \(h(x)=\sqrt{x-1}+2\).

Note that this transformation has changed the domain and range of the

function. This new graph has domain \([1,\infty)\) and range \([2,\infty)\).**Graphing Functions Using Reflections about the Axes**

Notice that the vertical reflection produces a new graph that is a mirror

image of the base or original graph about the y-axis. The horizontal reflection

produces a new graph that is a mirror image of the base or original graph

about the y-axis.**Reflections**

Given a function f(x), a new function g(x)=-f(x) is a vertical reflection of

the function f(x), sometimes called a reflection about the x-axis.

Given a function f(x), a new function g(x)=f(-x) is a horizontal reflection of

the function f(x), sometimes called a reflection about the y-axis.

Given a function, reflect the graph both vertically and horizontally

1. Multiply all outputs by -1 for a vertical reflection. The new graph

is a reflection of the original graph about the x-axis.

2. Multiply all outputs by -1 for a horizontal reflection. The new graph

is a reflection of the original graph about the y-axis.**Example 9**

Reflecting a Graph Horizontally and Vertically

Reflect the graph of \(s(t)=\sqrt{t}\) vertically and horizontally.

Solution:

Reflecting the graph vertically means that each output value will be reflected

over the horizontal t-axis.

Because each output value is the opposite of the original output value, we

can write: \(V(t)=-s(t)\).

Notice that his is an outside change, or vertical shift, that affects the

output s(t) values, so the negative sign belongs outside of the function.

Reflecting horizontally means that each input value will be reflected over

the vertical axis.

Because each input value is the opposite of the original input value, we

can write: \(H(t)=s(-t)\).

Notice that this is an inside change or horizontal change that affects the

input values, so the negative sign is on the inside of the function.

Note that these transformation can affect the domain and range of the

functions. While the original square root function has domain \([0,\infty)\) and

range \([0,\infty)\), the vertical reflection gives the V(t) function the range

\((-\infty.0])\) and the horizontal reflection gives the H(t) function the

domain \((-\infty,0]\).**Example 10**

Reflecting a Tabular Function Horizontally and Vertically

A function f(x) is given, create a table for the functions:

g(x)=-f(x)

h(x)=f(-x)

x 2 4 6 8

f(x)1 3 7 11

Solution:

For g(x), the negative sign outside the function indicates a vertical

reflection, so the x-values stay the same and each output value will be the

opposite of the original output value.

For h(x), the negative sign inside the function indicates a horizontal

reflection, so each input value will be the original input value and the h(x)

values stay the same as the f(x) values.**Example 11**

Applying a Learning Model Equation

A common model for learning has an equation similar to \(k(t)=2^{-2} + 1\),

where k is the percentage of mastery that can be achieved after t

practice sessions. This is a transformation of the function \(f(t)=2^t\).

Solution:

This equation combines three transformations into one equation.

1. A horizontal reflection:\(f(-t)=2^{-t}\)

2. A vertical reflection:\(-f(-t)=-2^{-t}\)

3. A vertical shift:\(-f(-t)=-2^{-t} + 1\)

We can sketch a graph by applying these transformations one at a

time to the original function. Let us follow two points through each of

the three transformations. We will choose the points (0,1) and (1,2).

1. First, we apply a horizontal reflection: (0,1)(-1,2)

2. Then, we apply a vertical reflection:(0,-1)(-1,-2)

3. Finally, we apply a vertical shift:(0,0)(-1,-1)

This means that the original points, (0,1) and (1,2) become (0,0) and (-1,-1)

after we apply the transformations.

As a model for learning, this function would be limited to a domain of

\(t \geq 0\), with a corresponding range of [0,1).**Determining Even and Odd Functions**

Some functions exhibit symmetry so that reflections result in the original

graph. For example, horizontally reflecting the toolkit functions

\(f(x)=x^2\) or \(f(x)=|x|\) will result in the original graph. We say

these types of graphs are symmetric about the y-axis. Functions whose graphs

are symmetric about the y-axis are called even functions. In the graphs

of \(f(x)=x^3\) or \(f(x)=1/x\) were reflected over both axes, the result

would be the original graph.

We say that these graphs are symmetric about the origin. A function with a

graph that is symmetric about the origin is called an odd function. A function

can be neither even nor odd if it does not exhibit symmetry. For example,

\(f(x)=2^x\) is neither even nor odd. Also, the only function that is both

even and odd is the constant function \(f(x)=0\).**Even and Odd Functions**

A function is called an even function if for every input of x:

\(f(x)=f(-x)\).

The graph of an even function is symmetric about the y-axis.

A function is called an odd function if for every input of x:

\(f(x)=-f(-x)\).

The graph of an odd function is symmetric about the origin.

Given the formula for a function, determine if the function is even, odd, or

neither.

1. Determine whether the function satisfies \(f(x)=f(-x)\). If it does, it is

even.

2. Determine whether the function satisfies \(f(x)=-f(-x)\). If it does, it

is odd.

3. If the function does not satisfy either rule, it is neither even nor odd.**Example 12**

Determining whether a Function is Even, odd, or neither

Is the function \(f(x)=x^3+2x\) even, odd, or neither?

Solution:

Without looking at a graph, we can determine whether the function is even or

odd by finding formulas for the reflections and determining if they

return us to the original function. Let us begin with the rule for even

functions.

\(f(-x)=(-x)^3 + 2(-x) = -x^3 - 2x\)

This does not return us to the original function, so this function is not

even. We can now test the rule for odd functions.

\(-f(-x)=-(-x^3-2x)=x^3+2x\)

Because \(-f(-x)=f(x)\), this is an odd function.

Consider the graph of f. Notice that the graph is symmetric about the

origin. For every point (x,y) on the graph, the corresponding

point(-x,-y) is also on the graph. For example, (1,3) is on the graph of f, and

the corresponding point (-1,-3) is also on the graph.**Graphing Functions Using Stretches and Compressions**

Adding a constant to the inputs of a function changed the position of a graph

with respect to the axes, but it did not affect the shape of a graph. We now

explore the effects of multiplying the inputs or outputs by some quantity.

We can transform the inside (input values) of a function or we can

transform the outside (output values) of a function. Each change has a

specific effect that can be seen graphically.**Vertical Stretches and Compressions**

When we multiply a function by a positive constant, we get a function whose

graph is stretched or compressed vertically in relation to the graph of

the original function. If the constant is greater than 1, we get a vertical

stretch. If the constant is between 0 and 1, we get a vertical compression. **Vertical Stretches and Compressions**

Given a function f(x), a new function g(x)=af(x), where a is a constant, is

a vertical stretch or vertical compression of the function f(x).

1. If a>1 then the graph will be stretched.

2. If 0<a<1, then the graph will be compressed.

3. If a<0, then there will be a combination of a vertical stretch with a

vertical reflection.

Given a function, graph its vertical stretch

1. Identify the value of a

2. Multiply all range values by a

3. If a>1, the graph is stretched by a factor of a

If 0<a(1, the graph is compressed by a factor of a

If a<0, the graph is either stretched or compressed and also reflected about

the x-axis.**Example 13**

Graphing a Vertical Stretch

A function P(t) models the population of fruit flies.

A scientist is comparing this population to another population, Q, whose

growth follows the same pattern, but is twice as large. Sketch a graph of

this population.

Solution:

Because the population is always twice as large, the new population's output

values are always twice the original function's output values.

If we choose four reference points, (0,1)(3,3)(6,2)(7,0), we will multiply all

of the outputs by 2.

The following shows where the new points for the new graph will be located.

This means that for any input t, the value of the function Q is twice the

value of the function P. Notice that the effect on the graph is a vertical

stretching of the graph, where every point doubles its distance from the

horizontal axis. The input values, t, stay the same while the output values

are twice as large as before.

Given a tabular function and assuming that the transformation is a

vertical stretch or compression, create a table for a vertical compression.

1. Determine the value of a

2. Multiply all of the output values by a**Example 14**

Finding a Vertical Compression of a Tabular Function

A function f is given, create a table for the function \(g(x)=\frac{1}{2}f(x)\).

x 2 4 6 8

f(x) 1 3 7 11

Solution:

The formula \(g(x)=\frac{1}{2}f(x)\) tells us that the output values of g are

half of the output values of f with the same inputs. For example, we know that

f(4)=3. Then

\(g(4)=\frac{1}{2}f(4)=\frac{1}{2}(3)=\frac{3}{2}\)

The result is that the function g(x) has been compressed vertically by

1/2. Each output value is divided in half, so the graph is half the

original height.**Example 15**

Recognizing a Vertical Stretch

The graph shown is a transformation of the toolkit function \(f(x)=x^3\).

Relate this new function g(x) to f(x), and then find a formula for g(x).

When trying to determine a vertical stretch or shift, it is helpful to look for

a point on the graph that is relatively clear. in this graph, it appears that

g(2)=2. With the basic cubic function at the same input, \(f(2)=2^3=8\).

based on that, it appears that the outputs of g are 1/4 the outputs of the

function f because g(2)=1/4 f92). From this we can safely conclude that

g9x)=1/4f(x). We can write a formula for g by using the definition of f.

\(g(x)=\frac{1}{4}f(x)=\frac{1}{4}x^3\)**Horizontal Stretches and Compressions**

Now we consider changes to the inside of the function. When we multiply a

function's input by a positive constant, we get a function whose graph is

stretched or compressed horizontally in relation to the graph of the original

function. If the constant is between 0 and 1, we get a horizontal stretch; if

the constant is greater than 1, we get a horizontal compression of the function.

Given a function y=f(x), the form y=f(bx) results in a horizontal stretch or

compression. Consider the function \(y=x^2\). The graph of \(y=(0.5x)^2\) is a

horizontal stretch of the graph of the function \(y=x^2\) by a factor of 1/2.**Horizontal Stretches and Compressions**

Given a function f(x), a new function g(x)=f(bx), where b is a constant, and is

a horizontal stretch or horizontal compression of the function f(x).

1. If b>1, then the graph will be compressed 1/b.

2. If 0<b<1, then the graph will be stretched by 1/b.

3. If b<0<1, then there will be combination of a horizontal stretch or

horizontal compression with a horizontal reflection.

Given a description of a function, sketch a horizontal compression or stretch.

1. Write a formula to represent the function

2. Set g(x)=f(bx) where b>1 for a compression or 0<b<1 for a stretch.**Example 16**

Graphing a Horizontal Compression

Suppose a scientist is comparing a population of fruit flies to a population

that progresses through its lifespan twice as fast as the original population.

In other words, this new population will progress in 1 hour the same amount as

the original population does in 2 hours, and in 2 hours, it will progress as

much as the original population does in 4 hours. Sketch a graph of this

population.

Symbolically, we could write:

\(R(1)=P(2)\)

\(R(2)=P(4)\)

\(R(t)=P(2t)\)

Note that the effect on the graph is a horizontal compression where all input

values are half of their original distance from the vertical axis.**Example 17**

Finding a Horizontal Stretch for a Tabular Function

A function f(x) is given, create a table for the function g(x)=f(1/2x).

x 2 4 6 8

f(x) 1 3 7 11

Solution:

The formula g(x)=f(1/2x) tells us that the output values for g are the same as

the output values for the function f at an input half the size. Notice that we

do not have enough information to determine g(2) because g(2)-f(1/2(2))=f(1),

and we do not have a value for f(1) in our table. Our input values to g will

need to be twice as large to get inputs for f that we can evaluate. For example,

we can determine g(4).

\(g(4)=f(1/2(4))=f(2)=1\)

Because each input value has been doubled, the result is that the function g(x)

has been stretched horizontally by a factor of 2.**Example 18**

Recognizing a Horizontal Compression on a Graph

Relate the function g(x) to f(x).

Solution:

The graph of g(x) looks like the graph of f(x) horizontally compressed. Because

f(x) ends at (6,4) and g(x) ends at (2,4), we can see that the x-values have

been compressed to 1/3, because 6(1/3)=2. We might also notice that g(2)=f(6)

and g(1)=f(3). Either way, we can describe this relationship as g(x)=f(3x). This

is a horizontal compression by 1/3.

Notice that the coefficient needed for a horizontal stretch or compression is

the reciprocal of the stretch or compression. So, to stretch the graph

horizontally by a scale factor of 4, we need a coefficient of 1/4 in our

function: f(1/4(x)). This means that the input values must be 4 times larger to

produce the same result, requiring the input to be larger, causing the

horizontal stretching.**Performing a Sequence of Transformations**

When combining transformations, it is very important to consider the order of

the transformations. For example, vertically shifting by 3 and then vertically

stretching by 2 does not create the same graph as vertically stretching by 2 and

then vertically shifting by 3, because when we shift first, both the original

function and the shift get stretched, while only the original function gets

stretched when we stretch first.

When we see an expression such as 2f(x)+3, which transformation should we start

with? The answer here follows nicely from the order of operations. Given the

output value of f(x), we first multiply by 2, causing the vertical stretch, and

then add 3, causing the vertical shift. In other words, multiplication before

addition.

Horizontal transformations are a little trickier to think about. When we write

g(x)=f(2x+3), for example, we have to think about how the inputs to the function

g relate to the inputs to the function f. Suppose we know f(7)=12. What input to

g would produce that output? In other words, what value of x will allow

g(x)=f(2x+3)=12? We would need 2x+3=7. To solve for x, we would first subtract

3, resulting in a horizontal shift, and then divide by 2, causing a horizontal

compression.

This format ends up being very difficult to work with, because it is usually

much easier to horizontally stretch a graph before shifting. We can work around

this by factoring inside the function.

\(f(bx+p)=f(b(x+\frac{p}{b}))\)

So:

\(f(x)=(2x+4)^2\)

We can factor out a 2

\(f(x)=(2(x+2))^2\)

Now we can more clearly observe a horizontal shift to the left 2 units and a

horizontal compression. Factoring in this way allows us to horizontally stretch

first and then shift horizontally.**Combining Transformations**

When combining vertical transformations written in the form af(x)+k, first

vertically stretch by a and then vertically shift by k.

When combining horizontal transformations written in the form f(bx-h), first

horizontally shift by h/b and then horizontally stretch by 1/b.

When combining horizontal transformations written in the form f(b(x-h)), first

horizontally stretch by 1/b and then horizontally shift by h.

Horizontal and vertical transformation are independent. It does not matter

whether horizontal or vertical transformations are performed first.**Example 19**

Finding a Triple Transformation of a Tabular Function

For the function f(x), create a table of values for the function g(x)=2f(3x)+1.

x 6 12 18 24

f(x) 10 14 15 17

Solution:

There are three steps to this transformation, and we will work from the inside

out. Starting with the horizontal transformations, f(3x) is a horizontal

compression by 1/3, which means we multiply each x-value by 1/3.

x 2 4 6 8

f(x) 10 14 15 17

Looking now to the vertical transformations, we start with the vertical

stretch, which will multiply the output values by 2. We apply this to the

previous transformation.

x 2 4 6 8

2f(3x) 20 28 30 34

Finally, we can apply the vertical shift, which will add 1 to all the output

values.

x 2 4 6 8

g(x)=2f(3x)+1 21 29 31 35**Example 20**

Finding a Triple Transformation of a Graph

Use the graph of f(x) to sketch a graph of k(x)=f(1/2x+1)-3

Solution:

To simplify, let us start by factoring out the inside of the function.

\(f(\frac({1}{2}x+1)-3=f(\frac{1}{2}(x+2)-3\)

By factoring the inside, we can first horizontally stretch by 2, as indicated by

the 1/2 on the inside of the function. Remember that twice the size of 0 is

still 0, so the point (0,2) remains at (0,2) while the point (2,0) will stretch

to (4,0).

Next, we horizontally shift left by 2 units, as indicated by x+2.

Last, we vertically shift down by 3 to complete our sketch, as indicated by the

-3 on the outside of the function.

**Quadratic Models**

When a mathematical model leads to a quadratic function, the properties of the

graph of the quadratic function can provide important information about the

model. We can use the quadratic function to determine the maximum or minimum

value of the function. The fact that the graph of a quadratic function has a

maximum or minimum value enables us to answer questions involving optimization

or finding the maximum or minimum values involving quadratic functions.**Solving Applied Problems**

In economics, revenue is defined as the amount of money received from the sale

of an item and is equal to the unit selling price of the item times the number

of units that were sold.

\(R=xp\)

In economics, the Law of Demand states that p and x are related. As one

increases, the other decreases. The equation that relates p and x is called the

demand equation.**Example 1**

The marketing department at Texas Instruments has found that, when certain

calculators are sold at a certain price, the number of calculators sold is given

by the demand equation:

\(x=21000-150p\)

1. Express the revenue as a function of price

2. What unit price should be used to maximize revenue

3. If this price is charged, what is the maximum revenue

4. How many units are sold at this price

5. Graph

Solution:

1. The revenue is \(R=xp\) where \(x=21000-150p\).

\((21000-150p)p= -150p^2+21000p\)

2. The function is a quadratic function with a=-150, b=21000, and c=0.

Because a<0, the vertex is the highest point on the parabola.

The revenue is therefore a maximum when the price is:

\(p=\frac{-b}{2a} = -\frac{21000}{2(-159)} = \frac{21000}{-300} = $70.00\)

3. The maximum revenue is:

\(R(70) = -150(70)^2 + 21000(70) = $735000\)

4. The number of calculators sold is given by the demand equation:

\(x=21000-150p\)

At a price of \(p=$70\),

\(x=21000-150(70) = 10500\)**Example 2**

Maximizing the Area Enclosed by a Fence

A farmer has 2000 yards of fence to enclose a rectangular field. What are the

dimensions of the rectangle that encloses the most area?

Solution:

The available fence represents the perimeter of the rectangle. If x is the

length and w is the width, then:

\(2x+2w=2000\)

The area of the rectangle is:

\(a=xw\)

To express area in terms of a single variable, we solve the equation for w and

substitute the result in \(a=xw\). Then area involves only the variable x. You

could also solve the equation for x and express area in terms of w alone.

\(2x + 2w = 2000\)

\(2w = 2000-2x\)

\(w = \frac{2000-2x}{2} = (1000-x)\)

So, the area is:

\(a = xw = x(1000-x) = -x^2 + 1000x\)

Now, area is a quadratic function of x.

\(a(x)=-x^2+1000x\)

When you graph this, you see a<0, so the vertex is a maximum point on the graph.

The maximum value occurs at:

\(x=-\frac{b}{2a} = -\frac{1000}{2(-1)} = 500\)

The maximum value of a is:

\(a(-\frac{b}{2a}) = a(500) = -500^2 + 1000(500) = -250000 + 500000 = 250000\)

The largest rectangle that can be enclosed by 2000 yards of fence has an area of

250000 square yards. Its dimensions are 500 by 500 yards.**Example 3**

Analyzing the motion of a projectile

A projectile is fired from a cliff 500 feet above the water at an inclination of

45 degrees to the horizontal, with a muzzle velocity of 400 feet per second. In

physics, it is established that the height of the projectile above the water is

given by:

\(h(x)=\frac{-32x^2}{(400)^2} + x + 500\)

X is the horizontal distance of the projectile from the base of the cliff.

1. Find the maximum height of the projectile

2. How far from the base of the cliff will the projectile strike the water?

Solution:

1. The height of the projectile is given by a quadratic function.

\(h(x)=\frac{-32x^2}{(400)^2}+x+500=\frac{-1}{5000}x^2+x+500\)

We are looking for the maximum value of h. Since a<0, the maximum value is

obtained at the vertex.

\(x=-\frac{b}{2a}=-\frac{1}{2(-1/5000)}=\frac{5000}{2}=2500\)

The maximum height of the projectile is:

\(h(2500)=\frac{-1}{5000}(2500)^2+2500+500=-1250+2500+500=1750ft\)

2. The projectile will strike the water when the height is zero. To find the

distance x traveled, we need to solve the equation:

\(h(x)=\frac{-1}{5000}x^2+x+500=0\)

We find the descriminant first.

\(b^2-4ac=1^2-4(\frac{-1}{5000})(500)=1.4\)

Then: \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm

\sqrt{1.4}}{2(-1/5000)}=5458ft\)

**Inequalities with Quadratic Functions**

In this section, we solve inequalities that involve quadratic functions. We will

accomplish this by using their graphs. For example, to solve the inequality

\(ax^2+bx+c>0\)

we graph the function \(f(x)=ax^2+bx+c\) and from the graph, determine where it

is above the x-axis and where f(x)>0. To solve the inequality, we graph the

function and determine where the graph is below the x-axis. If the inequality is

not strict, we include the x-intercepts in the solution.**Example 1**

Solving an Inequality

Solve the inequality \(x^2-4x-12\leq0\) and graph the solution set.

Solution:

We graph the function \(f(x)=x^2-4x-12\)

The intercepts are: x=6 and x=-2

the y-intercept is -12 with the x-intercepts at 6 and -2.

The vertex is at:

\(x=-\frac{b}{2a}=-\frac{-4}{2}=2\)

Since f(2)=-16, the vertex is (2,-16).

The graph is below the x-axis for -2<x<6. Since the original inequality is not

strict, we include the x-intercepts. The solution set is \({x|-2\leqx\leq6} or

[-2,6].**Example 2**

Solving an Inequality

Solve the inequality \(2x^2<x+10\) and graph the solution set.

Solution:

We arrange the inequality so that 0 is on the right side.

\(2x^2<x+10\)

\(2x^2-x-10<0\)

This inequality is equivalent to the one that we wish to solve.

Next, we graph the function \(f(x)=2x^2-x-10\). The intercepts are:

y=-10, x=-2, x=5/2

The vertex is at \(x=-\frac{b}{2a}=-\frac{-1}{4}=\frac{1}{4}\)

Since f(1/4)=-10.125, the vertex is at \((\frac{1}{4},-10.125)\)

The graph is below the x-axis between x=-2 and x=5/2. Since the inequality is

strict, the solution set is \({x|-2<x<\frac{5}{2}}\) or \((-2,\frac{5}{2})\).

**Polynomial Functions and Models**

**Polynomial Functions**

We can write linear functions as \(f(x)=mx+b\) and quadratic functions as

\(f(x)=ax^2+bx+c\). Each of these functions are polynomial functions. The

domain of a polynomial function is the set of all real numbers.

A polynomial function is a function whose rule is given by a polynomial in one

variable. The degree of a polynomial function is the largest power of x that

appears. The zero polynomial function \(f(x)=0\) is not assigned a degree.

Polynomial functions are among the simplest expressions in algebra. They are

easy to evaluate: only addition and multiplication are required. Because of

this, they are often used to approximate other, more complicated functions. **Example 1**

Identifying polynomial functions

Determine which of the following are polynomial functions. For those that are,

state the degree.

1. \(f(x)=2-3x^4\)

2. \(g(x)=\sqrt{x}\)

3. \(h(x)=\frac{x^2-2}{x^3-1}\)

4. \(f(x)=0\)

5. \(g(x)=8\)

6. \(h(x)=-2x^3(x-1)^2\)

Solution:

1. polynomial function of degree 4

2. not a polynomial function because the power is 1/2 which is not an integer

3. not a polynomial function. it is the ratio of two polynomials and the

polynomial in the denominator is of positive degree.

4. polynomial function of degree zero, it is not assigned a degree

5. nonzero constant function, which is a polynomial function of degree 0

6. polynomial function of degree 5

You will learn that the graph of every polynomial function is both smooth and

continuous. By smooth, we mean that the graph contains no sharp corners and by

continuous, we mean that the graph has no gaps or holes and can be drawn without

lifting pencil from paper.**Power Functions**

We begin the analysis of the graph of a polynomial function by discussing power

functions, a special kind of polynomial function. A power function of degree n

is a monomial of the form \(f(x)=ax^n\) where a is a real number, not zero, and

n>0.

Examples of power functions include:

\(f(x)=3x\) is degree 1

\(f(x)=-5x^2\) is degree 2

\(f(x)=8x^3\) is degree 3

\(f(x)=-5x^4\) is degree 4

The graph of a power function of degree 1, \(f(x)=ax\), is a straight line, with

slope a, that passes through the origin. The graph of a power function of degree

2, \(f(x)=ax^2\), is a parabola, with vertex at the origin, that opens up if a>0

and down if a<0.

If we know how to graph a power function of the form \(f(x)=x^n\), a compression

or stretch, and a reflection about the x-axis will enable us to obtain the graph

of \(g(x)=ax^n\). Consequently, we shall concentrate on graphing power functions

of the form \(f(x)=x^n\).

We begin with power functions of even degree of the form \(f(x)=x^n,n\geq2\) and

n even. The domain of f is the set of all real numbers, and the range is the set

of nonnegative real numbers. Such a power function is an even function, so its

graph is symmetric with respect to the y-axis. Its graph always contains the

origin and the points (-1,1) and (1,1).

If n=2, the graph is the familiar parabola \(y=x^2\) that opens up, with vertex

at the origin. If \(n\geq4\), the graph of \(f(x)=x^n\), n even, will be closer

to the x-axis than the parabola \(y=x^2\) if \(-1<x<1\) and farther from the

x-axis than the parabola \(y=x^2\) if \(x<-1\) or if \(x>1\). **Properties of Power Functions**

1. f is an even function, so its graph is symmetric with respect with respect to

the y-axis.

2. The domain is the set of all real numbers. The range is the set of

nonnegative real numbers.

3. The graph always contains the points (-1,1) and (1,1).

4. As the exponent n increases in magnitude, the graph becomes more vertical

when x<-1 or x>1. But for x near the origin, the graph tends to flatten out and lie closer to the x-axis.

Now we consider power functions of odd degree of the form \(f(x)=x^n,n\geq3\)

and n odd. The domain and range of f are the set of real numbers. Such a power

function is an odd function, so its graph is symmetric with respect to the

origin. Its graph always contains the points (-1,1) and (1,1).

The graph of \(f(x)=x^n\) when n=3 has been shown several times. If \(n\geq5\),

the graph of \(f(x)=x^n\), n odd, will be closer to the x-axis than that of

\(y=x^3\) if -1<x<1 and farther from the x-axis than that of \(y=x^3\) if x<-1

or if x>1.**Properties of Power Functions when x is Odd**

1. f is an odd function, so its graph is symmetric with respect to the origin

2. The domain and the range are the set of all real numbers.

3. The graph always contains the points (-1,-1) and (1,1).

4. As the exponent n increases in magnitude, the graph becomes more vertical

when x<-1 or x>1 but for x near the origin, the graph tends to flatten out

and lie closer to the x-axis.**Graph Polynomial Functions using Transformation**

The methods of shifting, compression, stretching, and reflection will enable us

to graph polynomial functions that are transformations of power functions.**Example 2**

Graph \(f(x)=1-x^5\)

Solution:

First, start with the usual function of \(y=x^5\)

Multiply by -1 to reflect across the x-axis.

This gives us \(y=-x^5\).

Now, add 1 to equation which will shift the graph up 1 unit.

\(y=-x^5+1\) which is the same as:

\(y=1-x^5\)**Example 3**

Graph \(f(x)=\frac{1}{2}(x-1)^4\)

Solution:

Take it step by step:

Start with the base function of \(y=x^4\)

Replace x by x-1 which will shift it right by 1 unit.

This gives us \(y=(x-1)^4\).

Now, multiple by 1/2 which gives the graph a compression by a factor of 1/2.

We now have \(y=\frac{1}{2}(x-1)^4\)**Finding the Real Zeros of a Polynomial Function**

The previous graph and problem shows a polynomial function with four

x-intercepts. Notice that at the x-intercepts the graph must either cross the

x-axis or touch the x-axis. Consequently, between consecutive x-intercepts the

graph is either above the x-axis or below the x-axis. We will make use of this

property of the graph of a polynomial function.

If a polynomial function f is factored completely, it is easy to solve the

equation \(f(x)=0\) using the zero-product property and locate the x-intercepts

of the graph. For example, if \(f(x)=(x-1)^2(x+3)\), then the solution of the

equation becomes:

\(f(x)=(x-1)^2(x+3)=0\) which factored give us: 1 and -3 as intercepts.

Based on this we can make the following observations.

If f is a function and r is a real number for which f(r)-0, then r is called a

real zero of f.

As a consequence, the following statements are equivalent.

1. r is a real zero of a polynomial function f

2. r is an x-intercept of the graph of f

3. x-r is a factor of f

So, the real zeros of a polynomial function are the x-intercepts of its graph

and they are found by solving the equation \(f(x)=0\).**Example 4**

Finding a Polynomial from its Zeros

Find a polynomial of degree 3 whose zeros are -3, 2, and 5.

Solution:

If r is a real zero of a polynomial f, then x-r is a factor of f. This means

that x-(-3)=x+3, x-2, and x-5 are factors of f. As a result, any polynomial of

the form \(f(x)=a(x+3)(x-2)(x-5)\) where a is any nonzero real number,

qualifies. The value of a causes a stretch, compression, or reflection, but does

not affect the x-intercepts.

If the same factor x-r occurs more than once, r is called a repeated or multiple

zero of f. More precisely, we have the following definition.

If (x-r)^m is a factor of a polynomial f and (x-r)^{m+1} is not a factor of f,

then r is called a zero of multiplicity m of f.**Example 5**

Identifying Zeros and their Multiplicities

For the polynomial \(f(x)=5(x-2)(x+3)^2(x-1/2)^4\)

Solution:

2 is a zero of multiplicity 1 because the exponent on the factor x-2 is 1

-3 is a zero of multiplicity 2 because the exponent on the factor x+3 is 2

1/2 is a zero of multiplicity 4 because the exponent on the factor x-1/2 is 4

Suppose that it is possible to factor completely a polynomial function and

locate all the x-intercepts of its graph(the real zeros of the function). These

x-intercepts then divide the x-axis into open intervals and, on each such

interval, the graph of the polynomial will be either above or below the x-axis.**Example 6**

Graphing a Polynomial Using its x-intercepts

For the polynomial \(f(x)=x^2(x-2)\)

1. Find the x and y intercepts of the graph of f

2. Use the x-intercepts to find the intervals on which the graph of f is above

the x-axis and the intervals on which the graph of f is below the x-axis.

3. Locate other points on the graph and connect all the points plotted with a

smooth continuous curve.

Solution:

1. The y-intercept is \(f(0)=0^2(0-2)=0\). The x-intercepts satisfy the

equation \(f(x)=x^2(x-2)=0\) from which we find \(x^2=0\) or \(x-2=0\) and

\(x=0\) or \(x=2\). The x-intercepts are 0 and 2

2. The two x-intercepts divide the x-axis into 3 intervals.

\((-\infty,0)(0,2)(2,\infty)\). Since the graph of f crosses or touches the x-axis

only at x=0 and x=2, it follows that the graph of f is either above the x-axis

f(x)>0 or below the x-axis f(x)<0 on each of these 3 intervals. To see where the

graph lies, we only need to pick a number in each interval, evaluate f there,

and see whether the value is positive(above the x-axis) or negative(below the

x-axis).

3. in constructing a table, we obtained three additional points on the

graph:(-1,-3)(1,-1)(3,9).

Since the graph of \(f(x)=x^2(x-2)\) is below the x-axis on both sides of 0, the

graph of f touches the x-axis at x=0, a zero of multiplicity 2. Since the graph

of f is below the x-axis for x<2 and above the graph for x>2, the graph of f

crosses the x-axis at x=2, a multiplicity of 1.

If r is a zero of even multiplicity the sign of f(x) does not change from one side to

the other side of r. The graph of f touches the x-axis at r.

If r is a zero of odd multiplicity, the sign of f(x) changes from one side to

the other side of r. The graph of f crosses the x-axis at r.**Behavior Near a Zero**

We have just learned how the multiplicity of a zero can be used to determine

whether the graph of a function touches or crosses the x-axis at the zero.

However, we can learn more about the behavior of the graph near its zeros than

just whether the graph crosses or touches the x-axis.

Consider the function\(f(x)=x^2(x-2)\). The zeros of f are 0 and 2. We can see

that the points on the graph of \(f(x)=x^2(x-2)\) and the points on the graph of

\(y=-2x^2\) are indistinguishable near x=0. So, \(y=-2x^2\) describes the

behavior of the graph of \(f(x)=x^2(x-2)\) near x=0.

But how did we know that the function \(f(x)=x^2(x-2)\) behaves like \(y=-2x^2\)

when x is close to 0? Where did \(y=-2x^2\) comes form? Because the zero, 0,

comes from the factor \(x^2\), we evaluate all factors in the function f at 0

with the exception of \(x^2\).

\(f(x)=x^2(x-2)\)

The factor \(x^2\) gives rise to the zero, so we keep the factor \(x^2\) and let

x=0 in the remaining factors to find the behavior near 0.

\(f(x)=x^2(0-2) = -2x^2\)

This tells us that the graph of \(f(x)=x^2(x-2)\) will behave like the graph of

\(y=-2x^2\) near x=0. Now let us discuss the behavior of \(f(x)=x^2(x-2)\) near

x=2, the other zero. Because the zero, 2, comes from the factor x-2, we

evaluate all factors of the function f at 2, with the exception of x-2.

\(f(x)=x^2(x-2) = 2^2(x-2) = 4(x-2)\)

The factor x-2 gives rise to the zero, so we keep the factor x-2 and let z=2 in

the remaining factors to find the behavior near 2.

So the graph of \(f(x)=x^2(x-2)\) will behave like the graph of \(y=4(x-2)\)

near x=2. We can see that the points on the graph of \(f(x)=x^2(x-2)\) and the

points on the graph of \(y=4(x-2)\) are indistinguishable near x=2. So

\(y=4(x-2)\), a line with slope 4, describes the behavior of the graph of

\(f(x)=x^2(x-2)\) near x=2.

By determining the multiplicity of a real zero, we determine that the graph

crosses or touches the x-axis at the zero. By determining the behavior of the

graph near the real zero, we determine how the graph touches or crosses the

x-axis.**Turning Points**

When looking at the previous graph, we cannot be sure how low the graph actually

goes between x=0 and x=2. but we do know that somewhere in the interval (0,2)

the graph of f must change direction (from decreasing to increasing). The points

at which a graph changes direction are called turning points. In calculus, such

points are called local maxima and local minima, and techniques for locating

them are given. So we shall not ask for the location of turning points in our

graphs. Instead, we will use the following result from calculus, which tells us

the maximum number of turning points that the graph of a polynomial function can

have.

If f is a polynomial function of degree n, then f has at most n-1 turning

points.

If the graph of a polynomial function f has n-1 turning points, the degree of f

is at least n.

For example, the graph of \(f(x)=x^2(x-2)\) is the graph of a polynomial of degree

3 and 3-1=2 turning points: one at (0,0) and the other somewhere between x=0 and

x=2.**Example 7**

Identifying the graph of a polynomial function

Which of the graphs could be the graph of a polynomial function? For those that

could, list the real zeros and state the least degree the polynomial can have.

For those that could not, say why not.

Solution:

1. This cannot be the graph of a polynomial function because of the gap that

occurs at x=-1. Remember, the graph of a polynomial function is continuous with

no gaps or holes.

2. This could be the graph of a polynomial function. It has 3 real zeros at -2,

1, and 2. Since the graph has 2 turning points, the degree of the polynomial

function must be at least 3.

3. This cannot be the graph of a polynomial function because of the cusp at x=1.

Remember, the graph of a polynomial function is smooth.

4. This could be the graph of a polynomial function. It has 2 real zeros at -2

and 1. Since the graph has 3 turning points, the degree of the polynomial

function is at least 4.

For very large values of x, either positive or negative, the graph of

\(f(x)=x^2(x-2)\) looks like the graph of \(y-=x^3\). To see why, we write f in

the form:

\(f(x)=x^2(x-2)=x^3-2x^2=x^3(1-\frac{2}{x})\)

Now, for large values of x, either positive or negative, the term 2/x is close to

0, so for large values of x:

\(f(x)=x^3-2x^2=x^3(1-\frac{2}{x})=x^3\)

The behavior of the graph of a function for large values of x, either positive

or negative, is referred to as its end behavior.**Example 8**

Identifying the graph of a polynomial function

Which of the graphs could be the graph of:

\(f(x)=x^4+5x^3+5x^2-5x-6\)

Solution:

The y-intercept of f is f(0)=-6 so we can eliminate any graph whose y-intercept

is positive. Being positive means the curve crosses the y-axis above 0.

We don't have any methods for finding the x-intercept of f, so we move on to

investigate the turning points of each graph. since f is of degree 4, the graph

of f has at most 3 turning points. So, we eliminate the graph that has 5 turning

points.

Now, we look at end behavior. For large values of x, the graph of f will behave

like the graph of \(y=x^4\). This eliminates the graph whose end behavior is

like the graph of \(y=-x^4\).

Only the graph left is the graph which has 3 turning points and behaves like

\(x^4\).**Summary of Polynomial Functions**

Degree of polynomial f: n

Maximum number of turning points: n-1

At a zero of even multiplicity: the graph of f touches the x axis

At a zero of odd multiplicity: the graph of f crosses the x-axis.

Between zeros, the graph of f is either above or below the x-axis

End behavior: For large |x|, the graph of f behaves like the graph of of

\(y=a_{n}x^{n}\).**Example 9**

Analyzing the Graph of a Polynomial Function

For the polynomial \(f(x)=x^3+x^2-12x\) :

1. Find the x and y-intercepts of the graph of f

2. Determine whether the graph crosses or touches the x-axis at each

x-intercept

3. End behavior: Find the power function that the graph of f resembles for large

values of |x|

4. Determine the maximum number of turning points on the graph of f

5. Determine the behavior of the graph of f near each x-intercept

6. Put all the information together to obtain the graph of f

Solution:

1. The y-intercept is 0 because \(f(0)=0^3 +0^2-12(0)\)

To find the x-intercepts, if any, we first factor f

\(x(x^2+x-12 = x(x+4)(x-3)\) so x=0, -4, and 3

2. Since each real zero is of multiplicity 1, the graph of f will cross the

x-axis at each x-intercept

3. End behavior: the graph of f resembles that of the power function \(y=x^3\)

for large values of |x|

4. The graph of f will contain at most two turning points because \(x^3\) is the

highest exponent

5. The three x-intercepts are -4, 0, and 3

Near -4: \(-4(x+4)(-4-3) = 28(x+4)\) A line with slope 28

Near 0: \(x(0+4)(0-3)=-12x\) A line with slope -12

Near 3: \(3(3+4)(x-3)=21(x-3)\) A line with slope 21

6. Put all of this together on a piece of paper then use a calculator to graph

and see if it is close.**Example 10**

Analyzing the Graph of a Polynomial Function

\(f(x)=x^2(x-4)(x+1)\)

Solution:

1. The y-intercept is 0 because f(0)=0

2. The x-intercepts are 0, 4, and -1

The intercept 0 is a multiplicity of 2 so the graph of f will touch the

x-axis 4 and -1 are zeros of multiplicity 1, so the graph of f will cross the x-axis at 4 and -1.

3. End behavior: the graph of f resembles that of the power function \(y=x^4\)

for large values of |x|

4. The graph of f will contain at most three turning points because of \(x^4\)

5. The three x-intercepts are -1, 0, and 4

Near -1: \(-1^2(-1-4)(x+1)=-5(x+1)\) A line with slope -5

Near 0: \(x^2(0-4)(0+1)=-4x^2\) A parabola opening down

Near 4: \(4^2(x-4)(4+1)=80(x-4)\) A line with slope 80

6. Graph it on paper then with a calculator**Summary for Analyzing the Graph of a Polynomial**

1. Find the y-intercept by letting x=0 and finding the value of f(0)

Find the x-intercepts, if any, by solving the equation f(x)=0

2. Determine whether the graph of f crosses or touches the x-axis at each

x-intercept

3. End behavior: find the power function that the graph of f resembles for large

values of |x|

4. Determine the maximum number of turning points on the graph of f

5. Determine the behavior of the graph of f near each x-intercept

6. Put all the information together to obtain the graph of f

For polynomial functions that have non-integer coefficients and for polynomials

that are not easily factored, we utilize a graphing utility early in the

analysis of the graph. This is because the amount of information that can be

obtained from algebraic analysis is limited.

**Properties of Rational Functions**

Ratios of integers are called rational numbers. Ratios of polynomial functions

are called rational functions. A rational function is of the form

\(r(x)=\frac{p(x)}{q(x)}\). P and Q are polynomial functions and q is not the

zero polynomial. The domain of a rational function is the set of all real

numbers except those for which the denominator is zero.**Example 1**

Find the domain of a rational function

1. The domain of \(r(x)=\frac{2x^2-4}{x+5}\) is the set of all real numbers

except -5. The domain is \({x|x\not=-5}\).

2. The domain of \(r(x)=\frac{1}{x^2-4}\) is the set of all real numbers except

-2 so the domain is \({x|x\not=-2,x\not=2}\).

3. The domain of \(r(x)=\frac{x^3}{x^2+1}\) is the set of all real numbers.

4. The domain of \(r(x)=\frac{3}{x^2+2}\) is the set of all real numbers.

5. The domain of \(r(x)=\frac{x^2-1}{x-1}\) is the set of all real numbers

except 1 so the domain is \({x|x\not=1}\).

If \(r(x)=\frac{p(x)}{q(x)}\) is a rational function and if p and q have no

common factors, then the rational function is said to be in lowest terms. For a

rational function in lowest terms, the real zeros, if any, of the numerator are

the x-intercepts of the graph will play a major role in the graph. The real

zeros of the denominator also play a major role in the graph. **Example 2**

Graphing \(y=\frac{1}{x^2}\)

Solution:

The domain is the set of all real numbers except 0. The graph has no y-intercept

because x can never equal 0. The graph has no x-intercept because the equation

\(y=0\) has no solution. Therefore, the graph will not cross either of the

coordinate axes. Because:

\(h(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = h(x)\)

It is an even function, so its graph is symmetric with respect to the y-axis. **Example 3**

Using transformations to graph a rational function

Graph the rational function \(r(x) = \frac{1}{(x-2)^2 + 1}\)

Solution:

First, we take note of the fact that the domain of r is the set of all real

numbers except x=2. To graph r, we start with the graph of \(y=\frac{1}{x^2}\).

Use a calculator to graph the function.**Asymptotes**

In the previous graph, notice that as the values of x become more negative as x

approaches negative infinity, the values approach 1. In fact, we can conclude

the following:

1. As x approaches negative infinity, the values of r(x) approach 1

2. As x approaches 2, the values of r(x) approach infinity

3. As x approaches infinity, the values of r(x) approach 1

This behavior of the graph is shown by the vertical line x=2 and the horizontal

line y=1. These lines are called asymptotes of the graph.

A horizontal asymptote, when it occurs, describes the end behavior of the graph

as x approaches infinity or as x approaches negative infinity. The graph of a

function may intersect a horizontal asymptote. A vertical asymptote, when it

occurs, describes the behavior of the graph when x is close to some number. The

graph of a function will never intersect a vertical asymptote.

There is a third possibility. If the value of a rational function approaches a

linear expression, then the line y=ax+b is an oblique asymptote. An oblique

asymptote, when it occurs, describes the end behavior of the graph. The graph of

a function may intersect an oblique asymptote.**Find the Vertical Asymptotes of a Rational Function**

The vertical asymptotes of a rational function, in lowest terms, are located at

the real zeros of the denominator. Suppose that r is a real zero of q, so x-r is

a factor of q. As x approaches r, the values of x-r approach 0, causing the

ratio to become unbounded, or head towards infinity. Based on the definition, we

conclude that the line x=r is a vertical asymptote. **Example 4**

Finding vertical asymptotes

Find the vertical asymptotes of the graph of each rational function

1. \(r(x)=\frac{x}{x^2-4}\)

2. \(r(x)=\frac{x+3}{x-1}\)

3. \(r(x)=\frac{x^2}{x^2+1}\)

4. \(r(x)=\frac{x^2-9}{x^2+4x-21}\)

Solution:

1. The function is in lowest terms and the zeros of the denominator are -2 and

The lines x=-2 and x=2 are the vertical asymptotes of the graph.

2. The function is in lowest terms and the only zero of the denominator is 1.

The line x=1 is the vertical asymptote of the graph.

3. The function is in lowest terms and the denominator has no real zeros,

because the equation has no real solutions. The graph has no vertical asymptotes

4. Factor to get it to lowest terms.

\(r(x)=\frac{x^2-9}{x^2+4x-21} = \frac{(x+3)(x-3)}{(x+7)(x-3)} =

\frac{x+3}{x+7}\) The only zero of the denominator in lowest terms is -7. The

line x=-7 is the only vertical asymptote of the graph.

Rational functions can have no vertical asymptotes, one vertical asymptote, or

multiple vertical asymptotes. However, the graph of a rational function will

never intersect any of its vertical asymptotes. **Find the Horizontal or Oblique Asymptotes of a Rational Function**

The procedure for finding horizontal and oblique asymptotes is somewhat more

involved. To find such asymptotes, we need to know how the values of a function

behave as x a[[roaches negative infinity or as x approaches infinity.

If a rational function is proper, that is, if the degree of the numerator is

less than the degree of the denominator, then as x approaches negative infinity

or as x approaches infinity the value of the function approaches 0.

Consequently, the line y=0(x-axis) is a horizontal asymptote of the graph. If a

rational function is proper, the line y=0 is a horizontal asymptote of its graph.**Example 5**

Finding Horizontal Asymptotes

Find the horizontal asymptotes, if any, of the graph of:

\(f(x)=\frac{x-12}{4x^2+x+1}\)

Solution:

Since the degree of the numerator is 1, it is less than the degree of the

denominator which is 2. So this is a proper function. That means that the line

y=0 is a horizontal asymptote of the graph.

To see why y=0 is a horizontal asymptote of the function, we need to investigate

the behavior as x approaches negative infinity and as x approaches infinity.

When |x| is unbounded, the numerator, which is x-12, can be approximated by the

power function y=x, while the denominator, which is \(4x^2+x+1\), can be

approximated by the power function \(y=4x^2\). We find:

\(f(x)=\frac{x-12}{4x^2+x+1} = \frac{x}{4x^2} = \frac{1}{4x}\) as it approaches

0. This shows that the line y=0 is a horizontal asymptote of the graph.

If a rational function is improper, or if the degree of the numerator is greater

than the degree of the denominator, we must use long division to write the

rational function as the sum of a polynomial plus a proper rational function.

1. If f(x)=b, a constant, the line y=b is a horizontal asymptote of the graph

2. If f(x)=ax+b, the line y=ax+b is an oblique asymptote of the graph

3. In all other cases, the graph of r approaches the graph of f, and there are

no horizontal or oblique asymptotes.**Example 6**

Finding Horizontal or Oblique Asymptotes

Find the horizontal or oblique asymptote of the graph of:

\(f(x)=\frac{3x^4-x^2}{x^3-x^2+1}\)

Solution:

Since the degree of the numerator, 4, is larger than the degree of the

denominator, 3, the rational function is improper. To find any horizontal or or

oblique asymptotes, use long division to get 3x+3, which is the oblique

asymptote.**Example 7**

Finding Horizontal or Oblique Asymptotes

Find asymptotes of:

\(f(x)=\frac{8x^2-x+2}{4x^2-1}\)

Solution:

Since the degree of the numerator, 2, equals the degree of the denominator,2,

the rational function is improper. To find any horizontal or oblique asymptotes,

use long division. We get y=2 as a horizontal asymptote of the graph.

We note that the quotient 2 obtained by long division is the quotient of the

leading coefficients of the numerator polynomial and the denominator polynomial.

This means that we can avoid the long division process for rational functions

whose numerator and denominator are of the same degree and conclude that the

quotient of the leading coefficients will give us the horizontal asymptote.**Example 8**

Finding the Horizontal or Oblique Asymptotes

\(f(x)=\frac{2x^5-x+2}{x^3-1}\)

Solution:

Since the degree of the denominator, 5, is larger than the degree of the

denominator, 3, the rational function is improper. To find any horizontal or

oblique asymptotes, use long division. \(2x^2-1\). Since this is not a linear

function, the function has no horizontal or oblique asymptotes.

1. If the degree of the numerator is less than the degree of the denominator, it

is a proper rational function, and the graph will have the horizontal asymptote

y=0.

2. If the degree of the numerator is greater than or equal to the degree of the

denominator, then it is an improper function. Use long division.

A. if the degree of the numerator equals the degree of the denominator, the

quotient obtained will be the number\(\frac{a_{n}}{b_{m}}\) and the line

\(y=\frac{a_{n}}{b_{m}}\) is a horizontal asymptote.

B. If the degree of the numerator is more than the degree of the

denominator, the quotient obtained is of the form ax+b and the line y=ax+b

is an oblique asymptote.

C. If the degree of the numerator is two or more than the degree of the

denominator, the quotient obtained is a polynomial of degree 2 or higher,

and the function has neither a horizontal nor an oblique asymptote. In this

case, for |x| unbounded, the graph will behave like the graph of the

quotient.

**Graphs of Rational Functions**

**Steps for Analyzing a Graph**

1. Factor the numerator and denominator and find the domain. If 0 is in the

domain, find the y-intercept, f(0), and plot it.

2. Write the function in lowest terms as \(\frac{p(x)}{q(x)}\) and find the real

zeros of the numerator. That is finding the real solution of the equation

p(x)=0. These are the x-intercepts of the graph. Determine the behavior of the

graph near each x-intercept, using the same procedure as for polynomial

functions. Plot each x-intercept and indicate the behavior of the graph.

3. With the function written in lowest terms, find the real zeros of the

equation q(x)=0. These determine the vertical asymptotes of the graph. Graph

each vertical asymptote using a dashed line.

4. Locate any horizontal or oblique asymptotes using the procedure given in the

previous section. Graph the asymptotes using a dashed line. Determine the points

at which the graph intersects these asymptotes. Plot any such points.

5. Using the real zeros of the numerator and the denominator of the given

equation, divide the x-axis into intervals and determine where the graph is

above the x-axis and where it is below the x-axis by choosing a number in each

interval and evaluating the function there. Plot the points found.

6. Analyze the behavior of the graph near each asymptote and indicate this

behavior in the graph.

7. Put all the information together to obtain the graph of the function given.**Example 1**

Analyze the graph of a rational function

\(r(x)=\frac{x-1}{x^2-4}\)

Solution:

1. We factor the numerator and denominator:

\(r(x)=\frac{x-1}{(x+2)(x-2)}\)

The domain is \({x|x \not=2, x \not =-2}\)

The y-intercept is \(r(0)=\frac{-1}{-4} = \frac{1}{4}\)

Plot the point \(0,\frac{1}{4}\)

2. Now that r is in lowest terms, the real zeros of the numerator satisfies the

equation x-1=0. The only x-intercept is 1.

Near 1: \(r(x)=\frac{x-1}{(x+2)(x-2)} = \frac{x-1}{(1+2)(1-2} =

\frac-{1}{3}(x-1)\)

Plot the point (1,0) and indicate a line with slope \(-\frac{1}{3}\) there.

3. R is in lowest terms. The real zeros of the denominator are the real

solutions of the equation (x+2)(x-2)=0, so -2 and 2. The graph of r has two

vertical asymptotes, the lines x=2 and x=-2. Graph each of these asymptotes

using dashed lines.

4. The degree of the numerator is less than the degree of the denominator, so r

is proper and the line y=0 (the x axis) is a horizontal asymptote of the graph.

Indicate this line by graphing y=0 using a dashed line. To determine if the

graph of r intersects the horizontal asymptote, we solve the equation r(x)=0.

\(\frac{x-1}{x^2-4}=0\)

\(x-1=0 >> x=1\)

The only solution is x=1. so the graph of r intersects the horizontal asymptote

at (1,0). We have already plotted this point.

5. The zero of the denominator, 1, and the zeros of the denominator, -2 and 2,

divide the x-axis into 4 intervals. Pick a point from each interval and solve

r(x). If the solution is negative then it is below the x-axis and if it is

positive then it is above the x-axis.

6. Next, we determine the behavior of the graph near the asymptotes.

A. Since the x-axis is a horizontal asymptote and the graph lies below the

x-axis for x<-2, we can sketch a portion of the graph by placing a small

arrow to the far left and under the x-axis.

B. Since the line x=-2 is a vertical asymptote and the graph lies below the

x-axis for x<-2, we continue by placing an arrow well below the x-axis and

approaching the line x=-2 on the left.

C. Since the graph is above the x-axis for -2<x<1 and x=-2 is a vertical

asymptote, the graph will continue on the right of x=-2 at the top.**Example 2**

Analyze the graph of a rational function

\(r(x)=\frac{x^2-1}{x}\)

Solution:

1. The domain of r is \({x|x\not=0}\). Because x cannot equal 0, there is not

y-intercept. Now factor r to obtain \(r(x)=\frac{(x+1)(x-1)}{x}\)

2. R is in lowest terms. Solving the equation r(x)=0, we find the graph has two

x-intercepts, -1 and 1.

Near -1: \(r(x)=\frac{(x+1)(x-1)}{x} = \frac{(x+1)(-1-1)}{-1} = 2(x+1)\)

Near 1: \(r(x)=\frac{(x+1)(x-1)}{x} = \frac{(1+1)(x-1)}{1} = 2(x-1)\)

Plot the point (-1,0) and indicate a line with slope 2 there. Plot the point

(1,0) and indicate a line with slope 2 there.

3. R is in lowest terms, so the graph of r has the line x=0 (the y-axis) as a

vertical asymptote. Graph x=0 using a dashed line.

4. The rational function r is improper, since the degree of the numerator, 2, is

larger than the degree of the denominator, 1. To find any horizontal or oblique

asymptotes, use long division. There is no solution so the graph of r does not

intersect the line y=x.

5. The zeros of the numerator are -1 and 1. The zero of the denominator is 0. We

use these values to divide the x-axis into four intervals. Pick a value in each

interval and solve for r(x). If the value is negative then the point is below

the x-axis. If the point is positive then the point is above the x-axis.

6. Since the graph of r is below the x-axis for x<-1 and is above the x-axis for

x>1, the graph of r will approach the line y=x. Since the graph of r is above

the x-axis for -1<x<0, the graph of r will approach the vertical asymptote x=0

at the top to the left of x=0. Since the graph of r is below the x-axis for

0<x<1, the graph of r will approach the vertical asymptote x=0 at the bottom

right of x=0.**Example 3**

Analyze the graph of a rational function

\(r(x)=\frac{x^4+1}{x^2}\)

Solution:

1. R is completely factored. The domain of r is \({x|x\not=0}\). There is no

y-intercept.

2. R is in lowest terms. Since \(x^4+1=0\) has no real solutions, there are no

x-intercepts.

3. R is in lowest terms, so x=0 (the y-axis) is a vertical asymptote of r. Graph

the line x=0 using dashes.

4. The rational function r is improper. To find any horizontal or oblique

asymptotes, use long division. We find the quotient to be \(x^2\), so the graph

has no horizontal or oblique asymptotes. However, the graph of r will approach

the graph of \(y=x^2\) as x approaches negative infinity and as x approaches

infinity. The graph of r does not intersect \(y=x^2\). Graph \(y=x^2\) using

dashes.

5. The numerator has no zeros, and the denominator has one zero at 0. We divide

the x-axis into the two intervals. Plot the points (-1,2) and (1,2).

6. Since the graph of r is above the x-axis and does not intersect \(y=x^2\), we

place arrows above \(y=x^2\). Also, since the graph of r is above the x-axis, it

will approach the vertical asymptote x=0 at the top to the left of x=0 and at

the top to the right of x=0.**Example 4**

Analyze the graph of a rational function

\(r(x)=\frac{3x^2-3x}{x^2+x-12}\)

Solution:

1. We factor r to get \(r(x)=\frac{3x(x-1)}{(x+4)(x-3)}\). The domain of r is

\({x|x\not=-4,x\not=3}\). The y-intercept is r(0)=0. Plot the point (0,0).

2. R is in lowest terms. Since the real solutions of the equation 3x(x-1)=0 are

x=0 and x=1, the graph has two x-intercepts, 0 and 1. We determine the behavior

of the graph of r near each x-intercept.

Near0:\(r(x)=\frac{3x(x-1)}{(x+4)(x-3)}=\frac{3x(0-1)}{(0+4)(0-3)}=\frac{1}{4}x\)

Near1:\(r(x)=\frac{3x(x-1)}{(x+4)(x-3)}=\frac{3(1)(x-1)}{(1+4)(1-3)}=-\frac{3}{10}(x-1)\)

Plot the point (0,0) and show a line with slope \(\frac{1}{4}\) there. Plot the

point (1,0) and show a line with slope \(-\frac{3}{10}\) there.

3. R is in lowest terms. Since the real solutions of the equation (x+4)(x-3)=0

are x=-4 and x=3, the graph of r has two vertical asymptotes, the lines x=-4 and

x=3. Plot these lines using dashes.

4. Since the degree of the numerator equals the degree of the denominator, the

graph has a horizontal asymptote. To find it, we either use long division or

form the quotient of the leading coefficient of the numerator, 3, and the

leading coefficient of the denominator, 1. The graph of r has the horizontal

asymptote y=3. To find out whether the graph of r intersects the asymptote, we

solve the equation r(x)=3.

\(r(x)=\frac{3x^2-3x}{x^2+x-12}=3\)

\(3x^2-3x=3x^2+3x-36\)

\(-6x=-36\)

\(x=6\)

The graph intersects the line y=3 only at x=6, and (6,3) is a point on the graph

of r. Plot the point (6,3) and the line y=3 using dashes.

5. The zeros of the numerator, 0 and 1, and the zeros of the denominator, -4 and

3, divide the x-axis into 5 intervals. Choose a number in each interval, solve

r(x) for each number chosen. If solution is positive then the point is above the

x-axis but if the solution is negative then the point is below the x-axis.

6. Since the graph of r is above the x-axis for x<-4 and only crosses the line

y=3 at (6,3), as x approaches negative infinity the graph of r will approach the

horizontal asymptote y=3 from above. The graph of r will approach the vertical

asymptote x=-4 at the top to the left of x=-4 and at the bottom to the right of

x=-4. The graph of r will approach the vertical asymptote x=3 at the bottom to

the left of x=3 and at the top to the right of x=3. We do not know whether the

graph of r crosses or touches the line y=3 at (6,3). To see whether the graph

crosses or touches the line y=3, we plot an additional point to the right of

(6,3) at x=6. Because (6,3) is the only point where the graph of r intersects

the asymptote y=3, the graph must approach the line y=3 from below as x

approaches infinity. **Example 5**

Analyze the graph of a rational function with a hole

Solution:

1. We factor r and obtain: \(r(x)=\frac{(2x-1)(x-2)}{x^2-4}\)

The domain of r is \({x|x\not=-2,x\not=2}\). The y-intercept is r(0)=-1/2. Plot

the point (0,-1/2).

2. In lowest terms, \(r(x)=\frac{2x-1}{x+2}\) \(x\not=-2\)

The graph has one x-intercept: 1/2.

Near 1/2: \(r(x)=\frac{2x-1}{x+2} = \frac{2x-1}{\frac{1}{2}+2} = 2/5(2x-1)\)

Plot the point (1/2,0) showing a line with slope 4/5.

3. Look at r in lowest terms. The graph has one vertical asymptote, x=-2, since

x+2 is the only factor of the denominator of r(x) in lowest terms. Remember,

though, the rational function is undefined at both x=2 and x=-2. Graph the line

x=-2 using dashes.

4. Since the degree of the numerator equals the degree of the denominator, the

graph has a horizontal asymptote. To find it, we use long division or form the

quotient of the leading coefficient of the numerator 2, and the leading

coefficient of the denominator, 1. The graph of r has the horizontal asymptote

y=2. Graph the line y=2 using dashes.

\(r(x)=\frac{2x-1}{x+2} = 2\)

\(2x-1=2(x+2)\)

\(2x-1=2x+4\)

\(-1=4\) Not a solution. The graph doe snot intersect the line y=2.

5. Look at the given expression for r. The zeros of the numerator and

denominator, -2,1/2, and 2, divide the x-axis into four intervals. Plot these

points.

6. We know the graph of r is above the x-axis for x<-2. We know the graph of r

does not intersect the asymptote y=2. Therefore, the graph of r will approach

y=2 from above as x approaches negative infinity and will approach the vertical

asymptote x=-2 at the top from the left. Since the graph of r is below the

x-axis for -2<x<1/2, the graph of r will approach x=-2 at the bottom from the

right. Finally, since the graph of r is above the x-axis for x>1/2 and does not

intersect the horizontal asymptote y=2, the graph of r will approach y=2 from below as x approaches infinity.

The real zeros of the denominator of a rational function give rise to either

vertical asymptotes or holes on the graph. **Example 6**

Constructing a rational function from its graph

Solution:

The numerator of a rational function in lowest terms determines the x-intercepts

of its graph. This graph has x-intercepts -2(even multiplicity, graph touches

the x-axis) and 5(odd multiplicity, graph crosses the x-axis). So one

possibility for the numerator is \(p(x)=(x+2)^2(x-5)\).

The denominator of a rational function in lowest terms determines the vertical

asymptotes of its graph. The vertical asymptotes of the graph are x=-5 and x=2.

Since r(x) approaches infinity to the left of x=-5 and r(x) approaches negative

infinity to the right of x=-5, then x+5 is a factor of odd multiplicity in q(x).

Also, because r(x) approaches negative infinity on both sides of x=2, then x-2

is a factor of even multiplicity in q(x). A possibility for the denominator is

\(q(x)=(x+5)(x-2)^2\). So far, we have:

\(r(x)=\frac{(x+2)^2(x-5)}{(x+5)(x-2)^2}\).

The horizontal asymptote of the graph is y=2, so we know that the degree of the

numerator must equal the degree of the denominator and the quotient of leading

coefficients must 2/1. This leads to:

\(r(x)=\frac{2(x+2)^2(x-5)}{(x+5)(x-2)^2}\)

**Polynomial Inequalities**

We solve inequalities that involve polynomials of degree 3 and higher, as well

as some that involve rational expressions. To solve such inequalities, we use

the information obtained previously about the graph of polynomial and rational

functions.

Suppose that the polynomial or rational inequality is of the form:

\(f(x)<0,f(x)>0,f(x)\leq0,f(x)\geq0\)

Locate the zeros of f if f is a polynomial function, and locate the zeros of the

numerator and the denominator if f is a rational function. If we use these zeros

to divide the real number line into intervals, we know that on each interval the

graph of f is either above the x-axis or below the x-axis. In other words, we

have found the solution of the inequality.**Steps for Solving Polynomial Inequalities**

1. Write the inequality so that a polynomial or rational expression is on the

left side and zero is on the right side. For rational expressions, be sure that

the left side is written as a single quotient and find its domain.

2. Determine the real numbers at which the expression on the left side equals

zero and, if the expression is rational, the real numbers at which the

expression on the left side is undefined.

3. Use the numbers found in step 2 to separate the real number line into

intervals.

4. Select a number in each interval and evaluate the function at the number.

A. If the value of f is positive, then f(x)>0 for all numbers in the interval

B. If the value of f is negative then f(x)<0 for all numbers in the interval

If the inequality is not strict, include the solutions of f(x)=0 in the solution

set.**Example 1**

Solving a Polynomial Inequality

\(x^4 \leq 4x^2\)

Solution:

Rearrange the inequality so that 0 is on the right side.

\(x^4 \leq 4x^2\)

\(x^4-4x^2 \leq 0\)

This inequality is equivalent to the one that we wish to solve.

Find the real zeros of \(f(x)=x^4-4x^2\) by solving its equation.

\(x^4-4x^2=0\)

\(x^2(x^2-4)=0\)

\(x^2(x+2)(x-2)=0\)

x=0 or x=-2 or x=2

We use these zeros to separate the real number line into four intervals.

select a number in each interval and evaluate \(f(x)=x^4-4x^2\) to determine if

f(x) is positive or negative.

We know that f(x)<0 for x in the intervals (-2,0) and (2,0), for all x such that

-2<x<0 or 0<x<2. however, because the original inequality is not strict, numbers

that satisfy the equation are also solution of the inequality. So, we include 0,

-2, and 2. The solution set of the given inequality is \({x|x-2 \leq x \leq

2}\). **Example 2**

Solving a polynomial inequality

\(x^4 > x\)

Solution:

Rearrange the inequality so that 0 is on the right side.

\(x^4-x > 0\)

This inequality is equivalent to the one that we wish to solve.

Find the real zeros of \(f(x)=x^4-x\) by solving \(x^4-x=0\)

\(x^4-x=0 \longrightarrow x(x^3-1)=0 \longrightarrow x(x-1)(x^2+x+1)=0\)

x=0 or x=1 The squared term is a quadratic and of course does not have any real

solutions.

We use the zeros 0 and 1 to separate the real number line into three intervals.

We choose a number in each interval and evaluate \(f(x)=x^4-x\) to determine if

the solution is positive or negative.

We know that f(x)>0 for all x in negative infinity to 0 and from 1 to infinity.

Because the original inequality is strict, the solution set of the given

inequality is \({x|x,0,x>1}\).**Example 3**

Solving a rational inequality

Solve \(\frac{(x+3)(2-x)}{(x-1)^2}>0\) and graph the solution set.

Solution:

The domain of the variable x is \({x|x\not=1}\). The inequality is already in a

form with 0 on the right side.

Let \(f(x)=\frac{(x+3)(2-x)}{(x-1)^2}\). The real zeros of the numerator of f

are -3 and 2. The real zero of the denominator is 1.

We use the zeros -3, 1 and 2 to separate the real number line into four

intervals.

Select a number in each interval and evaluate the number in

\(f(x)=\frac{(x+3)(2-x)}{(x-1)^2}\) to determine in f(x) is positive or

negative.

We know that f(x)>0 for all x in (-3,1) to (1,2). Because the original

inequality is strict, the solution set of the given inequality is

\({x|-3<x<2,x\not=1}\). Notice the hole at x=1 to indicate that 1 is to be

excluded.**Example 4**

Solving a rational inequality

\(\frac{4x+5}{x+2} \geq 3\)

Solution:

The domain of the variable x is \({x|x\not=-2}\). Rearrange the inequality so

that 0 is on the right side. Then express the left side as a single quotient.

\(\frac{4x+5}{x+2}-3 \geq 0\)

Multiply 3 by \(\frac{x+2}{x+2}\)

\(\frac{4x+5}{x+2}-3(\frac{x+2}{x+2} \geq 0 \)

Write as a single quotient

\(\frac{4x+5-3x+6}{x+2} \geq 0\)

Combine like terms

\(\frac{x-1}{x+2} \geq 0\)

The domain of the variable is \({x|x\not=-2}\).

Let \(f(x)=\frac{x-1}{x+2}\). The zero of the numerator is 1. The zero of the

denominator is -2.

We use the zeros -2, and 1 to separate the real number line into three

intervals.

Select a number in each interval and evaluate \(f(x)=\frac{4x+5}{x+2}-3\) to

determine if f(x) is positive or negative.

We know that f(x)>0 for all x from negative \((-\infty,-2)\) and \(1,\infty)\).

Because the original inequality is not strict, numbers that satisfy the equation

\(f(x)=\frac{x-1}{x+2}=0\) are also solutions of the inequality. Since

\(\frac{x-1}{x+2}=0\) only if x=1, we conclude that the solution set is

\({x|x<-2,x\geq1}\).

**Real Zeros of Polynomials**

In this section, we discuss the techniques that can be used to find the real

zeros of a polynomial function. Recall that if r is a real zero of a polynomial

function f then f(r)=0, r is an x-intercept of the graph of f, and r is a

solution of the equation f(x)=0.

For polynomial and rational functions, we have seen the importance of the real

zeros for graphing. In most cases, the real zeros of a polynomial function are

difficult to find using algebraic methods. No nice formulas like the quadratic

formula are available to help us find zeros for polynomials of degree 3 or

higher. Formulas do exist for solving any third or fourth degree polynomial

equation, but they are complicated to use. No general formulas exist for

polynomial equations of degree 5 or higher.**Remainder and Factor Theorems**

When we divide one polynomial by another, we obtain a quotient polynomial and a

remainder, the remainder being either the zero polynomial or a polynomial whose

degree is less than the degree of the divisor. To check our work, we do:

(quotient)(divisor) + remainder = dividend

This checking routine is the basis for a famous theorem called the division

algorithm for polynomials.**Remainder Theorem**

Let f be a polynomial function. If f(x) is divided by x-c, then the remainder is

f(c).**Example 1**

Using the remainder theorem

Find the remainder if \(f(x)=x^3-4x^2-5\) when divided by x-3 and x+2

Solution:

We could use long division, but who wants to do that, it is easier to use the

remainder theorem.

\(f(3)=(3)^3-4(3)^2-5 = 27-36-5 = 14\)

The remainder is 14

To find the remainder when f(x) is divided by x+2 = x-(-2) = -2

\(f(-2) = (-2)^3-4(-2)^2-5 = -8-16-5 = -29\)

The remainder is -29**Factor Theorem**

Let f be a polynomial function. Then x-c is a factor of f(x) if ands only if

f(c)=0. If f(c)=0, then x-c is a factor of f(x). If x-c is a factor of f(x),

then f(c)=0.

Suppose that f(c)=0. Then, we have \(f(x)=(x-c)q(x)\) for some polynomial q(x).

That is, x-c is a factor of f(x). Suppose that x-c is a factopr of f(x). Then

there is a polynomial function q such that \(f(x)=(x-c)q)x)\). Replacing x by c,

we find that \(f(c)=(c-c)q(c) = 0*q(c) = 0\). That completes the proof. One use

of the factor theorem is to determine whether a polynomial has a particular

factor.**Example 2**

Use the factor theorem to determine whether a function has a factor

\(f(x)=2x^3-x^2+2x-3\)

For x-1 and x+3

Solution:

That factor theorem states that if f(c)=0 then x-c is a factor

Because x-1 is of the form x-c with c=1, we find the value of f(1). We choose to

use substitution.

\(f(1) = 2(1)^3-(1)^2+2(1)-3 = 2-1+2-3 = 0\)

By the factor theorem, x-1 is a factor of x.

To test the factor x+3, we first need to write it in the form of x-c. Since

x+3=x-(-3), we find the value of f(-3). Because f(-3) =-72 and not 0, we

conclude from the factor theorem that x-(-3) or x+3 is not a factor of f(x).**Number of Real Zeros**

The next theorem concerns the number of real zeros that a polynomial function

may have. In counting the zeros of a polynomial, we count each zero as many

times as its multiplicity.

A polynomial function cannot have more real zeros than its degree.

The proof is based on the factor theorem. If r is a real zero of a polynomial

function f, then f(r)=0 and so, x-r is a factor of f(x). Each real zero

corresponds to a factor of degree 1. Because f cannot have more first degree

factors than its degree.

Descarte's rule of signs provides information about the number and location of

the real zeros of a polynomial function written in standard form. It requires

that we count the number of variations in the sign of the coefficients of f(x)

and f(-x). For example, the following polynomial function has two variations in

the signs of the coefficients.

\(f(x)=3x^7+4x^4+3x^2-2x-1 = -3x^7+0x^6+0x^5+4x^4+0x^3+3x^2-2x-1\)

Notice that we ignored the zero coefficients in \(0x^6,0x^5\) and \(0x^3\) in

counting the number of variations in the sign of f(x). Replacing x by -x, we

get:

\(f(-x)=-3(-x)^7+4(-x)^4+3(-x)^2-2(-x)-1 = 3x^7+4x^4+3x^2+2x-1\)

This has one variation in sign.

Descarte's Rule of Signs

Let f denote a polynomial function written in standard form.

The number of positive real zeros of f either equals the number of variations in

the sign of the nonzero coefficients of f(x) or else equals that number less an

even integer.

The number of negative real zeros of f either equals the number of variations in

the sign of the nonzero coefficients of f(-x) or else equals that number less an

even integer.**Example 3**

Using the number of real zeros theorem and the rule of signs.

Discuss the real zeros of \(f(x)=3x^6-4x^4+3x^3+2x^2-x-3\)

Solution:

Because the polynomial is of degree 6, by the real zeros theorem there are at

most six real zeros. Since there are three variations in the sign of the nonzero

coefficients of f(x), by the rule of signs, we expect either three or one

positive real zero. To continue, look at f(-x).

\(f(-x)=3x^6-4x^4-3x^3+2x^2+x-3\)

There are three variations in sign, so we expect either three or one negative

real zeros. Equivalently, we know now that the graph of f has either three or one

positive x-intercepts and three or one negative x-intercepts.**Rational Zeros Theorem**

The rational zeros theorem provides information about the rational zeros of a

polynomial with integer coefficients.

Let f be a polynomial function of degree 1 or higher of the form:

\(f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...a_{1}x+a_{0}\)

Each coefficient is an integer. If p/q is in lowest terms, it is a rational zero

of f, then p must be a factor of \(a_o\), and q must be a factor of \(a_n\).**Example 4**

Listing potential rational zeros

\(f(x)=2x^3+11x^2-7x-6\)

Solution:

Because f has integer coefficients, we may use the rational zero theorem. First,

we list all the integers p that are factors of the constant term a0=-6 and all

the integers q that are factors of the leading coefficient ay=2.

\(p:....\pm1,\pm2,\pm3,\pm6\)

\(q:....\pm1,\pm2\)

Now, we form all possible ratios p/q

\(\frac{p}{q}:....\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\)

If f has a rational zero, it will be found in this list, which contains 12

possibilities.

Be sure that you understand what the rational zeros theorem says. For a

polynomial with integer coefficients, if there is a rational zero, it is one of

those listed. It may be the case that the function does not have any rational

zeros.

Long division, synthetic division, or substitution can be used to test each

potential rational zero to determine whether it is indeed a zero. To make the

work easier, integers are usually tested first.**Example 5**

Finding the rational zeros of a polynomial function

\(f(x)=2x^3+11x^2-7x-6\)

Write f in factored form

Solution:

We gather all the information that we can about the zeros.

1. There are at most 3 zeros

2. By the rule of signs, there is one positive real zero. Also, because

\(f(-x)=-2x^3+11x^2+7x-6\)

There are two negative zeros or no negative zeros.

3. Now we use the list of potential rational zeros obtained in the previous

example.

\(\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\)

We choose to test the potential rational zero using substitution.

\(f(1)=2(1)^3+11(1)^2-7(1)-6 = 2+11-7-6 = 0\)

Since f(1)=0, 1 is a zero and x-1 is a factor of f. We can use long division or

synthetic division to factor f.

\(f(x)=2x^3+11x^2-7x-6 = (x-1)(2x^2+13x+6)\)

Now, any solution of the equation \(2x^2+13x+6=0\) will be a zero of f. Because

of this, we call the equation a depressed equation of f. Since the degree of the

depressed equation of f is less than that of the original polynomial, it is

easier to work with the depressed equation to find the zeros of f.

4. The depressed equation \(2x^2+13x+6=0\) is a quadratic equation with

discriminant \(b^2-4ac = 169-48 = 121 > 0\). The equation has two real

solutions, which can be found by factoring:

\(2x^2+13x+6 = (2x+1)(x+6) = 0\)

\(2x+1 = 0\) and \(x+6 = 0\)

So, \(x=-\frac{1}{2}\) or \(x = -6\)

The zeros of f are \(-6,-\frac{1}{2},1\)

We know that \(f(x)=2x^3+11x^2-7x-6\) can be written as

\(f(x)=(x-1)(2x^2+13x+6)\). Because \(2x^2+13x+6 = (2x+1)(x+6)\), we write f in

factored form as: \(f(x)=(x-1)(2x+1)(x+6)\)

Notice that the three zeros of f are found in this example are among those given

in the list of potential zeros. **Find the Real Zeros of a Polynomial Function**

1. Use the degree of the polynomial to determine the maximum number of real

zeros.

2. Use the rule of signs to determine the possible number of positive zeros and

negative zeros.

3. If the polynomial has integer coefficients, use the rational zeros theorem to

identify those rational numbers that potentially could be zeros.

Use substitution, synthetic division, or long division to test each potential

rational zero. Each time that a zero(and thus a factor) is found, repeat step 3

on the depressed equation.

4. In attempting to find the zeros, remember to use the factoring techniques

that you already know.**Example 6**

Finding the real zeros of a polynomial function

Find the real zeros of \(f(x)=x^5-5x^4+12x^3-24x^2+32x-16\)

Write f in factored form.

Solution:

We gather all the information that we can about the zeros.

There are at most 5 real zeros because of \(x^5\).

By the rule of signs, there are 5, three, or one positive zero.

There are no negative zeros because \(f(x)=-x^5-5x^4-12x^3-24x^2-32x-16\) has no

variation in sign.

Because the leading coefficient \(a_{5}=1\) and there are no negative zeros, the

potential rational zeros are the integers 1,2,4,8, and 16, the positive factors

of the constant term, 16. We test the potential rational zero 1 first, using

synthetic division.

We get a remainder of f(1)=0, so 1 is a zero and x-1 is a factor of f. Using the

entries in the bottom row of the synthetic division, we can begin to factor f.

\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x^4-4x^3+8x^2-16x+16)\)

We now work with the first depressed equation:

\(q1(x)=x^4-4x^3+8x^2-16x+16=0\)

The potential rational zeros of q1 are still 1,2,4,8, and 16. We test 1 first

since it may be a repeated zero of f.

Since the remainder is 5, 1 is not a repeated zero. We try 2 next:

The remainder is f(2)=0 so 2 is a zero and x-2 is a factor of f. Again using the

bottom row, we find:

\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x-2)(x^3-2x^2+4x-8)\)

The remaining zeros satisfy the new depressed equation:

\(q2(x)=x^3-2x^2+4x-8=0\)

Notice that q2(x) can be factored using grouping.

Since \(x^2+4=0\) has no real solutions, the real zeros of f are 1 and 2, the

latter being a zero of multiplicity 2. The factored form of f is:

\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x-2)^2(x^2+4)\)**Example 7**

Solving a polynomial equation

Find the real solutions of the equation:

\(x^5-5x^4+12x^3-24x^2+32x-16=0\)

Solution:

The real solutions of this equation are the real zeros of the polynomial

function \(f(x)=x^5-5x^4+12x^3-24x^2+32x-16\)

Using the result from previous example, the real zeros of f are 1 and 2. These

are the real solutions of the equation:

\(x^5-5x^4+12x^3-24x^2+32x-16=0\)

In example 6, the quadratic factor \(x^2+4\) that appears in the factored form

of f is called irreducible, because the polynomial \(x^2+4\) cannot be factored

over the real numbers. In general, we say that a quadratic factor \(ax^2+bx+c\)

is irreducible if it cannot be factored over the real numbers, that is, if it is

prime over the real numbers.

Refer to examples 5 and 6. The polynomial function of example 5 has three real

zeros, and its factored form contains three linear factors. The polynomial

function of example 6 has two distinct real zeros, and its factored form

contains two distinct linear factors and one irreducible quadratic factor.

Every polynomial function with real coefficients can be uniquely factored into a

product of linear factors and/or irreducible quadratic factors.

We shall prove this result and we shall draw several additional conclusions

about the zeros of a polynomial function. One conclusion is worth noting. If a

polynomial with real coefficients is of odd degree, it must contain at least one

linear factor. This means it must have at least one real zero.

A polynomial function with real coefficients of odd degree has at least one real

zero.**Theorem for Bounds on Zeros**

The search for the real zeros of a polynomial function can be reduced somewhat

if bounds on the zeros are found. A number M is a bound on the zeros of a

polynomial if every zero lies between -M and M. That is, M is a bound on the

zeros of a polynomial if \(-M \leq\text{ any real zero of f }\leq M\)**Example 8**

Using the theorem for finding bounds on zeros

Find a bound on the real zeros of each polynomial

1. \(f(x)=x^5+3x^3-9x^2+5\)

2. \(g(x)=4x^5-2x^3+2x^2+1\)

Solution:

1. The leading coefficient of f is 1.

\(f(x)=x^5+3x^3-9x^2+5 = 0,3,-9,0 = 5\)

The smaller of the two numbers, 10, is the bound. Every real zero of f lies

between -10 and 10.

2. First, we write g so that it is the product of a constant times a polynomial

whose leading coefficient is 1.

\(g(x)=4x^5-2x^3+2x^2+1 = 4(x^5-\frac{1}{2}x^3+\frac{1}{2}x^2+\frac{1}{4}\)

Next, we evaluate the two expressions in (4) with

\(a_4=0,a_3=-\frac{1}{2},a_2=\frac{1}{2},a_1=0\text{ and }a_0=\frac{1}{4}\)

The smaller of the two numbers, \(\frac{5}{4}\) is the bound. Every real zero

of g lies between \(-\frac{5}{4}\text{ and }\frac{5}{4}\)**Intermediate Value Theorem**

It is based on the fact that the graph of a polynomial function is continuous,

that is, it contains no holes or gaps. Let f denote a polynomial function. If

a<b and if f(a) and f(b) are of opposite sign, there is at least one real zero

of f between a and b.**Example 9**

Using the intermediate value theorem to locate real zeros

Show that \(f(x)=x^5-x^3-1\) has a real zero between 1 and 2.

Solution:

We evaluate f at 1 and 2

\(f(1)=-1\)

\(f(2)=23\)

Because f(1)<0 and f(2)>0, it follows from the intermediate value theorem that f

has a zero between 1 and 2.

Let us look at the polynomial f of example 9 more closely. Based on the rule of

signs, f has exactly one positive real zero. Based on the rational zeros

theorem, 1 is the only potential positive rational zero. Since f(1) does not

equal 0, we conclude that the zero between 1 and 2 is irrational. We can use the

intermediate value theorem to approximate it.

1. Find two consecutive integers a and a+1 such that f has a zero between them.

2. Divide the interval into 10 equal subintervals.

3. Evaluate f at each endpoint of the subintervals until the intermediate value

theorem applies. This endpoint then contains a zero.

4. Repeat the process until the desired accuracy is achieved.**Example 10**

Approximating the real zeros of a polynomial function

Find the positive zero of:

\(f(x)=x^5-x^3-1\) to two decimal places.

Solution:

From example 9, we know that the positive zero is between 1 and 2. We divide the

interval [1,2] into 10 equal subintervals.

[1,1.1][1.1,1.2][1.2,[1.3][1.3,1.4][1.4,1.5][1.5,1.6][1.6,1.7][1.7,1.8][1.9,1.9][1.9,2].

Now we find the value of f at each endpoint until the intermediate value theorem

applies.

\(f(x)=x^5-x^3-1\)

f(1.0)=-1 f(1.2)=-0.23968 f(1.1)=-0.72049 f(1.3)=0.51593

We can stop here and conclude that the zero is between 1.2 and 1.3. now we

divide the interval[1.2,1.3] into 10 equal subintervals and proceed to evaluate

f at each endpoint. We conclude that the zero lies between 1.23 and 1.24 and

correct to two decimal places, the zero is 1.23.

**Complex Zeros**

**Complex Zeros**

A variable in the complex number system is referred to as a complex variable. A

complex number r is called a complex zero of f if f(r)=0.

We have learned that some quadratic equations have no real solutions, but that

in the complex number system every quadratic equation has a solution, either

real or complex.

Every complex polynomial function f(x) of degree \(n \geq 1\) has at least one

complex zero.

Let \(f(x)=a_n^n+a_{n-1}x^{n-1}+...+a_1x+a_0\)

By the fundamental theorem of algebra, f has at least one zero, r1. Then, by the

factor theorem, x-r1 is a factor and \(f(x)=(x-r_1)q_1(x)\) where where q1(x) is

a complex polynomial of degree n-1 whose leading coefficient is an. Repeating

this argument n times, we arrive at \(f(x)=(x-r_1)(x-r_2)...(x-r_n)q_n(x)\)

where \(q_n(x)\) is a complex polynomial of degree n-n=0 whose leading

coefficient is \(a_n\). That is, \(q_n(x)=a_nx^n=a_n\), and so

\(f(x)=a_n(x-r_1)(x-r_2)...(x-r_n)\). We conclude that every complex polynomial

function f(x) of degree \(n \geq 1\) has exactly n zeros. **Conjugate Pairs Theorem**

We can use the fundamental theorem of algebra to obtain valuable information

about the complex zeros of polynomials whose coefficients are real numbers.

In other words, for polynomials whose coefficients are real numbers, the complex

zeros occur in conjugate pairs. This result should not be all that surprising

since the complex zeros of a quadratic function occurred in conjugate pairs.

The importance of this result should be clear. Once we know that, say, \(3+4i\)

is a zero of a polynomial with real coefficients, then we know that \(3-4i\) is

also a zero. This result has an important corollary. A polynomial f of odd

degree with real coefficients has at least one real zero.

For example, the polynomial \(f(x)=x^5-3x^4+4x^3-5\) has at least one zero that

is a real number, since f is of degree 5 (odd) and has real coefficients. **Example 1**

Using the conjugate pairs theorem

A polynomial f of degree 5 whose coefficients are real numbers has the zeros 1,

5i, and 1+i. Find the remaining two zeros.

Solution:

Since f has coefficients that are real numbers, complex zeros appear as

conjugate pairs. It follows that -5i, the conjugate of 5i, and 1-i, the

conjugate of 1+i, are the two remaining zeros.**Example 2**

Finding a polynomial function whose zeros are given

Find a polynomial function f of degree 4 whose coefficients are real numbers

that has the zeros 1, 1, and -4+i.

Solution:

since -4+i is a zero, by the conjugate pairs theorem, -4-i must also be a zero

of f. Because of the factor theorem, if f(c)=0, then x-c is a factor of of f(x).

So we can now write f as:

\(f(x)=a(x-1)(x-1)[x-(-4+i)][x-(-4-i)]\) where a is any real number.

\(f(x)=a(x^2-2x+1)[x^2-(-4+i)x-(-4-i)x+(-4+i)(-4-i)]\)

\(a(x^2-2x+1)(x^2+4x-ix+4x+ix+16+4i-4i-i^2)\)

\(a(x^2-2x+1)(x^2+8x+17)\)

\(a(x^4+8x^3+17x^2-2x^3-16x^2-34x+x^2+8x+17)\)

\(a(x^4+6x^3+2x62-26x+17)\)

Every polynomial function with real coefficients can be uniquely factored over

the real numbers into a product of linear factors and/or irreducible quadratic

factors.

This second degree polynomial has real coefficients and is irreducible over the

real numbers. So, the factors of f are either linear or irreducible quadratic

factors. **Example 3**

Finding the complex zeros of a polynomial

\(f(x)=3x^4+5x^3+25x^2+45x-18\)

Solution:

1. The degree of f is 4, so f will have four complex zeros.

2. The rule of signs provides information about the real zeros. For this

polynomial, there is one positive real zero. There are three or one negative

real zeros because \(f(-x)=3x^4-5x^3+25x^2-45x-18\) has three variations in

sign.

3. The rational zeros theorem provides information about the potential rational

zeros of polynomials with integer coefficients. For this polynomial, which has

integer coefficients, the potential rational zeros are:

\(\pm 1/3, \pm 2/3, \pm 1, \pm 2, \pm 6, \pm 9, \pm 18\)

Since f(-2)=0, then -2 is a zero and x+2 is a factor of f. the depressed

equation is:

\(3x^3-x^2+27x-9=0\)

Factor by grouping

\(x^2(3x-1)+9(3x-1)=0\)

Factor out the common factor 3x-1.

\((x^2+9)(3x-1)=0\)

\(x^2+9=0\)

\(3x-1=0\)

\(x^2=-9\)

\(x=1/3\)

\(x=-3i,3i,1/3\)

So the complex zeros are \(-3i.3i.-2,1/3\)

Factored form of f is:

\(f(x)=3x^4+5x^3+25x^2=45x-18 = 3(x+3i)(x-3i)(x+2)(x-1/3)\)