# Rates of Change in Precalculus

These are my notes on rates of change in precalculus.

**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.