Question 2: How do you calculate the average rate of change from a function?

When a function is given by a formula, we modify the definition of average rate of

change:

Average Rate of Change

The average rate of change of ( )f x with respect to x from

x a to x b is defined as

Average rate of change of ( ) ( )with respect to over ,

f f b f ax a b b a

Close examination of this definition reveals that it is simply the same definition as

before, but with

Change in ( ) ( )

Change in

f f b f a

x b a

so that we can write

Average rate of change of Change in ( ) ( )with respect to over , Change in

f f f b f ax a b x b a

Instead of using the data values to calculate the changes in the difference quotient, we

use the function’s formula to get the values in the numerator of the difference quotient.

In the context of the graph of a function, the average rate of change of a function can be

visualized as the slope of a line that passes through two points on the function. This

line, called a secant line, can be drawn on a graph of a function so that we can quantify

the value of the slope of the line.

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Figure 1 - The average rate of change of f(x) with respect to x over [a, b] is equal to the slope of a line of a secant line.

A secant line passing through the points , ( )a f a and , ( )b f b has a vertical rise of

( ) ( )f b f a and a horizontal run of b a . The slope of between the points on the secant

line is

( ) ( )f f b f a

x b a

Example 4 Find the Average Rate of Change from a Function’s

Formula

The annual sales (in millions of dollars) at Apple from 2001 through 2010

can be modeled by

0.284( ) 3405.120 tS t e

where t is the number of years since 2000. Find the average rate of

change of sales with respect to time over the period from 2001 to 2010.

(Modeled from Apple Annual Reports)

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Solution Using the definition of the average rate of change,

Average rate of change of (10) (1)with respect to over 1,10 10 1

S S St

The function values in the numerator are computed from the function’s

formula,

0.284 1

0.284 10

(1) 3405.120 4523.474

(10) 3405.120 58281.236

S e

S e

This leads to the average rate of change,

Average rate of change of 58281.236 4523.474with respect to over 1,10 10 1

53757.762 million dollars

9 years

5973.085 million dollars per year

S

t

Each decimal is written to three decimal places since the decimals in

the original function were written to three decimal places. This average

rate tells us that the sales increased by an average of about 5973.085

million dollars or $5,973,085,000 in each year from 2001 through 2010.

On a graph of the sales function ( )S t , the average rate of change of sales from 1t to

10t may be visualized as the slope of the line connecting the points at 1, (1)S and

10, (10)S .

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Figure 2 - The slope of the line connecting the sales at t = 1 and t = 10 is the same as the average rate of change between these points.

In calculus, one difference quotient in particular comes up very often.

The average rate of change of ( )f x with respect to x from

x a to x a h is

Average rate of change of ( ) ( )

with respect to over ,

f f a h f ax a a h h

This is the same definition we have been using with functions given by formulas, but

with b a h . The value of h is the horizontal separation of the two points on the secant

line. This difference quotient will be very important in defining the instantaneous rate of

change in Section 11.2.

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Figure 3 - An alternate description of average rate of change.

Example 5 Find the Difference Quotient

Let ( )f x be defined by

2( ) 3f x x x

a. Find and simplify the difference quotient

(2 ) (2)f h f

h

Solution This difference quotient is

( ) ( )f a h f a

h

with 2a . Find the function values in the numerator of the difference

quotient. The value (2)f is found by replacing x with 2 in the function,

2(2) 2 3 2 2f

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Similarly, (2 )f h is found by replacing x with 2 h ,

 

2

2

2

2

(2 ) 2 3 2

4 4 3 2

4 4 6 3

2

f h h h

h h h

h h h

h h

 

Substitute these function values into the difference quotient to yield

b. Evaluate the difference quotient in part a when 0.25h .

Solution The expression for the difference quotient in part a, 1h ,

makes it easy to evaluate the difference quotient for any value of h. It

enables us to find the average rate of change of ( )f x with respect to x

from 2x and any other x value.

Set 0.25h to give

(2 0.25) (2)0.25 1 1.25

1

f f

This value is the average rate of change of ( )f x with respect to x from

2x to 2 0.25x .

Square 2 h

Remove the parentheses

Simplify

Remove the parentheses

Simplify the numerator

Factor the numerator

Reduce

2

2

2

2 2(2 ) (2)

2 2

1

1

h hf h f

h h

h h

h

h h

h

h h

h

h

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Care must be taken to calculate ( )f a h in the difference quotient. The most common

mistake in simplifying the function value ( )f a h is to assume it is the sum of the values

of the function. For instance, in the previous example (2 ) (2) ( )f h f f h . The powers

on the factors correspond to multiplication. They must be worked out when calculating

( )f a h .

Example 6 Find the Difference Quotient

Find and simplify the difference quotient

(3 ) (3)g h g

h

for the function

3( )g t t

Solution To find the difference quotient, we must evaluate ( )g t at 3 and

3 h . The value 3(3) 3 27g is easy to calculate. However (3 )g h is

more challenging:

33 3g h h

To simplify the expression on the right, we need to cube the quantity

3 h

2

2

2 2 3

3 2

3 3 3 3

3 9 6

9 3 6 3 3

27 9 18 6 3

9 27 27

g h h h h

h h h

h h h h h

h h h h h

h h h

Multiply 3 3h h

Multiply 3 h times each term

Remove the parentheses

Simplify

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Now we can put the function values into the difference quotient and

simplify:

3 2

3 2

2

2

9 27 27 27(3 ) (3)

9 27

9 27

9 27

h h hg h g

h h

h h h

h

h h h

h

h h

Simplify the numerator

Factor

Reduce

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