## Reading: Average Rate of Change

Since functions represent how an output quantity varies with an input quantity, it is natural to ask about the rate at which the values of the function are changing.

For example, the function C(t) below gives the average cost, in dollars, of a gallon of gasoline t years after 2000.

 t 2 3 4 5 6 7 8 9 C(t) 1.47 1.69 1.94 2.3 2.51 2.64 3.01 2.14

If we were interested in how the gas prices had changed between 2002 and 2009, we could compute that the cost per gallon had increased from $1.47 to$2.14, an increase of $0.67. While this is interesting, it might be more useful to look at how much the price changed per year. You are probably noticing that the price didn’t change the same amount each year, so we would be finding the average rate of change over a specified amount of time. The gas price increased by$0.67 from 2002 to 2009, over 7 years, for an average of $\displaystyle \frac{{{0.67}}}{{{7}{y}{e}{a}{r}{s}}}approx{0.096}$ dollars per year. On average, the price of gas increased by about 9.6 cents each year.

## Rate of Change

rate of change describes how the output quantity changes in relation to the input quantity. The units on a rate of change are “output units per input units.”

Some other examples of rates of change would be quantities like:

• A population of rats increases by 40 rats per week
• A barista earns $9 per hour (dollars per hour) • A farmer plants 60,000 onions per acre • A car can drive 27 miles per gallon • A population of grey whales decreases by 8 whales per year • The amount of money in your college account decreases by$4,000 per quarter

## Average Rate of Change

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.

Average rate of change =

### Example 1

Using the cost-of-gas function from earlier, find the average rate of change between 2007 and 2009

From the table, in 2007 the cost of gas was $2.64. In 2009 the cost was$2.14.

The input (years) has changed by 2. The output has changed by $2.14 –$2.64 = –0.50. The average rate of change is then
= –0.25 dollars per year

### Try it Now 1

Using the same cost-of-gas function, find the average rate of change between 2003 and 2008

Notice that in the last example the change of output was
negative since the output value of the function had decreased. Correspondingly, the average rate of change is negative.

### Example 2

Given the function g(t) shown here, find the average rate of change on the interval [0, 3].

At t = 0, the graph shows

At t = 3, the graph shows

The output has changed by 3 while the input has changed by 3, giving an average rate of change of:

### Example 3

On a road trip, after picking up your friend who lives 10 miles away, you decide to record your distance from home over time. Find your 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

You traveled 282 miles in 6 hours, for an average speed ofHere, your average speed is the average rate of change.

We can more formally state the average rate of change calculation using function notation.

## Average Rate of Change using Function Notation

Given a function f(x), the average rate of change on the interval [a, b] is

Average rate of change = $\displaystyle\frac{{\text{Change of Output}}}{{\text{Change of Input}}}=\frac{{\Delta{y}}}{{\Delta{x}}}=\frac{{{y}_{{2}}-{y}_{{1}}}}{{{x}_{{2}}-{x}_{{1}}}}$

### Example 4

Compute the average rate of change of $\displaystyle{f{(x)}={x}^{{2}}-\frac{{1}}{{x}}$ on the interval [2, 4]

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

$\displaystyle{f(2)}={2}^{{2}}-\frac{{1}}{{2}}={4}-\frac{{1}}{{4}}=\frac{{7}}{{2}}$

$\displaystyle{f(4)}={4}^{{2}}-\frac{{1}}{{4}}={16}-\frac{{1}}{{4}}=\frac{{63}}{{4}}$

Now computing the average rate of change

Average rate of change =
$\displaystyle\frac{{f(4)-f(2)}}{{{4}-{2}}}=\frac{{\frac{{63}}{{4}}-\frac{{7}}{{2}}}}{{{4}-{2}}}=\frac{{\frac{{49}}{{4}}}}{{2}}=\frac{{49}}{{8}}$

## Try it Now 2

Find the average rate of change of $\displaystyle{f(x)}={x}-{2}\sqrt{{x}}$ on the interval [1, 9]

### Example 5

The magnetic force F, measured in Newtons, between two magnets is related to the distance between the magnets d, in centimeters, by the formula $\displaystyle{F(d)}=\frac{{2}}{{{d}^{{2}}}}$. Find the average rate of change of force if the distance between the magnets is increased from 2 cm to 6 cm.

We are computing the average rate of change of $\displaystyle{F(d)}=\frac{{2}}{{{d}^{{2}}}}$on the interval [2, 6]

Average rate of change $\displaystyle=\frac{{{F(6)}-{F(2)}}}{{{6}-{2}}}$

 Evaluating the Function $\displaystyle\frac{{{F(6)}-{F(2)}}}{{{6}-{2}}}$ $\displaystyle\frac{{\frac{{2}}{{6}^{{2}}}-\frac{{2}}{{2}^{{2}}}}}{{{6}-{2}}}$ Simplifying $\displaystyle\frac{{\frac{{2}}{{36}}-\frac{{2}}{{4}}}}{{4}}$ Combing the numerator terms $\displaystyle\frac{{\frac{{-{16}}}{{36}}}}{{4}}$ Simplifying further $\displaystyle\frac{{-{1}}}{{9}}$ Newtons per centimeter

This tells us the magnetic force decreases, on average, by
$\displaystyle\frac{{-{1}}}{{9}}$ Newtons per centimeter over this interval.

### Example 6

Find the average rate of change of g(t) = t2 + 3t + 1on the interval [0, a]. Your answer will be an expression involving
a.

 Evaluating the Function $\displaystyle\frac{{{g(a)} -{g(0)}}}{{{a}-{0}}}$ $\displaystyle\frac{{{({a}^{{2}}+{3}{a}+{1})}-{({0}^{{2}}-{3}{({0})}+{1})}}}{{{a}-{0}}}$ Simplifying $\displaystyle\frac{{{a}^{{2}}+{3}{a}+{1}-{1}}}{{a}}$ Simplifying further, and factoring $\displaystyle\frac{{{a}{({a}+{3})}}}{{a}}$ Canceling the common factor $\displaystyle{a}$ $\displaystyle{a}+{3}$

This result tells us the average rate of change 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.

### Try it Now 3

Find the average rate of change of f(x) = x3 + 2 on the interval [a,a + h].

# Important Topics of This Section

• Rate of Change
• Average Rate of Change
• Calculating Average Rate of Change using Function Notation

1. $\displaystyle\frac{{{3.01}-{1.69}}}{{{5}\text{years}}}=\frac{{{1.32}}}{{{5}{\text{years}}}=$0.264 dollars per year.
2. Average rate of change $\displaystyle=\frac{{{f{{({9})}}}-{f{{({1})}}}}}{{{9}-{1}}}=\frac{{{({9}-{2}\sqrt{{9}})}-{({1}-{2}\sqrt{{1}})}}}{{{9}-{1}}}=\frac{{{3}-{(-{1})}}}{{{9}-{1}}}=\frac{{4}}{{8}}=\frac{{1}}{{2}}$
3. $\displaystyle\frac{{{f{{({a}+{h})}}}-{f{{({a})}}}}}{{{({a}+{h})}-{a}}}=\frac{{{({({a}+{h})}^{{3}}+{2})}-{({a}^{{3}}+{2})}}}{{h}}=\frac{{{a}^{{3}}+{3}{a}^{{2}}{h}+{3}{a}{h}^{{2}}+{h}^{{3}}+{2}-{a}^{{3}}-{2}}}{{h}}=\frac{{{3}{a}^{{2}}{h}+{3}{a}{h}^{{2}}+{h}^{{3}}}}{{h}}=\frac{{{h}{({3}{a}^{{2}}+{3}{a}{h}+{h}^{{2}})}}}{{h}}={3}{a}^{{2}}+{3}{a}{h}+{h}^{{2}}$