math.wiki
Functions › What a function is › Graphs of functions › Features of a graph › Average rate of change › Average rate of change from a formula
Open on the map
Algebra 1 · F.IF.B.6

Average rate of change from a formula

You have the rule for a quantity. How fast did it change, on average, between two input values?

You need
Evaluating a function at an expressionAlgebra 1
Unlocks
The difference quotientPrecalculusMaximum, minimum and range of a quadratic functionAlgebra 1
Where you are 4 in Average rate of change
Functions13
What a function is5
Graphs of functions5
Features of a graph10
Average rate of change4
Average rate of change from a formula
Average rate of change from a table or graphRates at an instant from shrinking intervalsThe difference quotient

When you reach for it

A rule gives a quantity’s value for any input: the height of a ball t seconds after it is thrown, a savings balance after t years, the users of an app after m months. Often the question is not “how much?” but “how fast?”: how many meters per second the ball climbed between two moments, how many dollars per year the balance gained. That number is the average rate of change over the stretch of inputs, the Rate of change measured from one end of the stretch to the other.

With the rule in hand you need no table and no graph. You evaluate the function at the two ends, then one subtraction and one division finish the job. When the values come from a table or a graph instead, the same quotient is read off two rows or two points, as in Average rate of change from a table or graph.

For a straight-line rule the answer is the same on every interval. For anything else it changes with the interval, so the interval is part of the question: “the average rate of change of h on [0, 1]”, never just “the rate of change of h”.

The procedure

The average rate of change of a function f on the interval [a, b], where a \ne b, is the change in output divided by the change in input:

\frac{f(b) - f(a)}{b - a}

The numerator is how much the output changed between the two ends; the denominator is the width of the interval. Dividing turns a total change into a change per unit of input, so the units are output units per input unit: meters per second, dollars per year, users per month.

Why call it an average? Suppose the output had changed at one steady rate r the whole way. Starting from f(a) and gaining r for each of the b - a units of input, it would arrive at f(a) + r(b - a). Setting that equal to f(b) and solving for r gives exactly the quotient above. So the average rate of change is the one steady rate that would carry the function from its starting value to its finishing value, whatever it actually did in between.

The order of the ends must match top and bottom. Reversing both differences, \displaystyle \frac{f(a) - f(b)}{a - b}, gives the same number, since the minus signs on top and bottom cancel; reversing only one flips the sign and turns a rise into a fall.

Take g(x) = x^2 + 4x on [1, 3]. The values at the ends are g(1) = 1 + 4 = 5 and g(3) = 9 + 12 = 21, so the average rate of change is

\begin{aligned} \frac{g(3) - g(1)}{3 - 1} &= \frac{21 - 5}{2} = 8. \end{aligned}

As a check on the meaning: start at 5, gain a steady 8 per unit for 2 units, and you reach 5 + 16 = 21. On [-4, 0], g(-4) = g(0) = 0, so the average rate is \displaystyle \frac{0 - 0}{4} = 0, although the output dipped below zero in between.

For a linear rule L(x) = mx + k the quotient is the slope m on every interval. The constant k cancels in the subtraction, and b - a cancels in the division:

\begin{aligned} \frac{L(b) - L(a)}{b - a} &= \frac{(mb + k) - (ma + k)}{b - a} \\ &= \frac{m(b - a)}{b - a} = m. \end{aligned}

That is what makes a line a line. Linear functions are exactly the ones with a single rate of change; every other function needs an interval named.

Worked examples

Example 1

The same rate on every interval

Find the average rate of change of k(x) = 13 - 6x on [-1, 2] and on [3, 10].

Solution

On [-1, 2], the values at the ends are k(-1) = 13 + 6 = 19 and k(2) = 13 - 12 = 1. The width is 2 - (-1) = 3, so

\begin{aligned} \frac{k(2) - k(-1)}{2 - (-1)} &= \frac{1 - 19}{3} \\ &= \frac{-18}{3} = -6. \end{aligned}

On [3, 10], k(3) = 13 - 18 = -5 and k(10) = 13 - 60 = -47. The width is 7, so

\begin{aligned} \frac{k(10) - k(3)}{10 - 3} &= \frac{-47 - (-5)}{7} \\ &= \frac{-42}{7} = -6. \end{aligned}

Both give -6, the coefficient of x, as the procedure showed for every linear rule: the output falls by 6 for each unit the input rises, wherever you look.

Example 2

A ball thrown upward

A ball is thrown upward from a platform, and its height in meters t seconds later is

h(t) = -5t^2 + 25t + 2.

Find the average rate of change of h on [0, 1], [1, 2], [3, 4] and [0, 5], and say what each means.

Solution

First the heights at the whole seconds, each by substituting into the rule:

t012345
h(t)2223232222

For example h(1) = -5 + 25 + 2 = 22 and h(4) = -80 + 100 + 2 = 22. The first three intervals are one second wide and the last is five:

\begin{aligned} \frac{h(1) - h(0)}{1 - 0} &= \frac{22 - 2}{1} = 20 \\ \frac{h(2) - h(1)}{2 - 1} &= \frac{32 - 22}{1} = 10 \\ \frac{h(4) - h(3)}{4 - 3} &= \frac{22 - 32}{1} = -10 \\ \frac{h(5) - h(0)}{5 - 0} &= \frac{2 - 2}{5} = 0 \end{aligned}

The units are meters per second. In the first second the ball climbs at an average 20 meters per second; in the next only 10, because it slows as it rises. From t = 3 to t = 4 the rate is negative: the ball has passed its highest point and is coming down. Over the whole flight the average is 0, because the ball ends at the height it started from.

t
h
(0,\,2)
(1,\,22)
(2,\,32)
(3,\,32)
(4,\,22)
(5,\,2)
Figure 1The ball of Example 2. It rises 20 meters in the first second, only 10 in the next, then falls back the same way.
Example 3

Users doubling each month

A new phone app has 300 users at launch, and in a made-up model its users double each month, so after m months it has U(m) = 300 \cdot 2^m users. Find the average rate of change of U on [0, 1], [1, 2], [2, 3] and [0, 3].

Solution

The values are U(0) = 300, U(1) = 600, U(2) = 1200 and U(3) = 2400. The one-month intervals give

\begin{aligned} \frac{600 - 300}{1} &= 300, \\ \frac{1200 - 600}{1} &= 600, \\ \frac{2400 - 1200}{1} &= 1200 \end{aligned}

users per month, and the whole three months give

\begin{aligned} \frac{U(3) - U(0)}{3 - 0} &= \frac{2400 - 300}{3} \\ &= \frac{2100}{3} = 700 \end{aligned}

users per month.

The rate doubles from each month to the next. Evaluating the rule at an expression shows why: on any one-month interval [m, m + 1] the gain is

\begin{aligned} U(m + 1) - U(m) &= 300 \cdot 2^{m + 1} - 300 \cdot 2^m \\ &= 300 \cdot 2^m (2 - 1) = 300 \cdot 2^m, \end{aligned}

the number of users at the start of that month: doubling means gaining as many as you already have. The three-month average of 700 is the rate in no single month.

Example 4

Interest on a growing balance

Hester puts 2000 dollars in a savings account that adds 4\% of the balance each year, so after t years the balance is B(t) = 2000(1.04)^t dollars. Find the average rate of change of B on [0, 1] and on [10, 11], to the nearest cent.

Solution

On [0, 1], B(0) = 2000 and B(1) = 2000(1.04) = 2080, so the average rate is 80 dollars per year.

On [10, 11] the powers need a calculator:

\begin{aligned} B(10) &= 2000(1.04)^{10} \approx 2960.49, \\ B(11) &= 2000(1.04)^{11} \approx 3078.91, \end{aligned}

so the average rate is

\begin{aligned} \frac{B(11) - B(10)}{11 - 10} &\approx \frac{3078.91 - 2960.49}{1} \\ &= 118.42 \end{aligned}

dollars per year.

Each year adds 4\% of whatever is there, and 4\% of 2960.49 dollars is about 118.42 dollars. So, as with the app, the rate keeps rising on later intervals of the same width.

Example 5

Which grows faster from 0 to 4?

Compare p(x) = 2^x and q(x) = 4x + 1. Which has the greater average rate of change on [0, 4]? Which on [4, 6]?

Solution

Both start at p(0) = q(0) = 1. At the other end p(4) = 16 and q(4) = 17. So on [0, 4]

\begin{aligned} \frac{p(4) - p(0)}{4} &= \frac{15}{4} = 3.75, \\ \frac{q(4) - q(0)}{4} &= \frac{16}{4} = 4. \end{aligned}

Over the first four units the line gains more. On [4, 6], p(6) = 64 and q(6) = 25, so

\begin{aligned} \frac{p(6) - p(4)}{2} &= \frac{64 - 16}{2} = 24, \\ \frac{q(6) - q(4)}{2} &= \frac{25 - 17}{2} = 4. \end{aligned}

Now the doubling rule gains six times as fast.

So “which grows faster” has no answer until the interval is named. The line’s rate is always 4, while the rate of p on [x, x + 1] is 2^{x + 1} - 2^x = 2^x, below 4 while x < 2 and above it once x > 2.

x
y
x = 4
y = 2^x
y = 4x + 1
Figure 2Example 5. On [0, 4] the line gains more; past x = 4 the doubling rule pulls away.
Example 6

Every one-second interval at once

For the ball of Example 2, h(t) = -5t^2 + 25t + 2, find the average rate of change on [t, t + 1] as an expression in t. Use it to find the one-second interval on which the average rate is 0.

Solution

The ends of the interval are now letters, so this is evaluating the rule at an expression. Put t + 1 in for the input, expand, and collect:

\begin{aligned} h(t + 1) &= -5(t + 1)^2 + 25(t + 1) + 2 \\ &= -5t^2 - 10t - 5 + 25t + 25 + 2 \\ &= -5t^2 + 15t + 22. \end{aligned}

Subtract h(t); the -5t^2 terms cancel:

\begin{aligned} h(t + 1) - h(t) &= (-5t^2 + 15t + 22) \\ &\quad - (-5t^2 + 25t + 2) \\ &= -10t + 20. \end{aligned}

The width is (t + 1) - t = 1, so the average rate on [t, t + 1] is -10t + 20 meters per second.

This one expression holds every one-second answer of Example 2: t = 0 gives 20, t = 1 gives 10, t = 3 gives -10, and t = 4 gives -20, matching h(5) - h(4) = 2 - 22. Each interval that starts a second later has an average rate 10 meters per second lower.

The rate is 0 when -10t + 20 = 0, that is t = 2, so the interval is [2, 3], the one Example 2 skipped. Check: h(2) = -20 + 50 + 2 = 32 and h(3) = -45 + 75 + 2 = 32. The interval is centered on t = 2.5, the moment the ball is highest, and the equal heights either side of the peak are what make the average zero.

Boundaries

  • The two ends must differ. When a = b the denominator is 0 and the quotient is undefined. A rate at a single instant needs a further idea, below.
  • The function must be defined at both ends. For \displaystyle f(x) = \sqrt{x} there is no average rate on [-4, 0], because f(-4) is not a real number. If the rule breaks inside the interval, the quotient may still compute but describes no single journey: \displaystyle f(x) = \frac{1}{x} on [-1, 1] gives \displaystyle \frac{1 - (-1)}{2} = 1, yet the graph has no point at x = 0.
  • The average sees only the ends. Two functions that agree at a and at b have the same average rate on [a, b], whatever each does in between.
  • The rate depends on the input unit. The same growth is 80 dollars per year or about 6.67 dollars per month, so the unit belongs with the answer.

Where this goes next

Example 6 found the average rate on an interval of width 1 with a letter at its left end. Letting the width be a letter too, [x, x + h], gives the difference quotient, one expression for the average rate on every interval of every width. Shrinking the width toward zero turns an average over an interval into a rate at an instant, such as the ball’s speed at exactly t = 1; in calculus that is called the derivative.

The zero average rate on [2, 3] was no accident: for a quadratic, an interval whose ends have equal outputs is centered on the vertex, one way to locate the maximum or minimum of a quadratic.

Exercises

Linear and quadratic rules

  1. Find the average rate of change of f(x) = 2x + 11 on [1, 6].
  2. Find the average rate of change of \displaystyle g(x) = 8 - \frac{1}{3}x on [-3, 6].
  3. Find the average rate of change of s(x) = x^2 + 2 on [1, 4].
  4. Find the average rate of change of s(x) = x^2 + 2 on [-4, -1].
  5. Find the average rate of change of s(x) = x^2 + 2 on [-2, 2].
  6. Find the average rate of change of w(x) = 3x^2 - x on [2, 5].

Exponential rules

  1. For f(x) = 2^x, find the average rate of change on [0, 2], on [2, 4] and on [4, 6]. How does each compare with the one before?
  2. Find the average rate of change of v(x) = 3^x on [-1, 1].
  3. Find the average rate of change of \displaystyle d(x) = 64\left(\frac{1}{2}\right)^x on [0, 3] and on [3, 6].

In context

  1. For the ball of Example 2, h(t) = -5t^2 + 25t + 2, find the average rate of change on [0.5, 1.5] and on [3.5, 4.5], with units.
  2. Feodora throws a ball whose height in meters after t seconds is k(t) = -5t^2 + 15t + 2. Find its average rate of change on [0, 1] and on [1, 3], and say what the signs tell you.
  3. Isambard’s savings balance after t years is B(t) = 500(1.06)^t dollars. Find the average rate of change on [0, 5] and on [5, 10], to the nearest cent.
  4. A made-up app has N(m) = 150 \cdot 2^m users after m months. Find the average rate of change over the first four months, [0, 4], and over the fifth month, [4, 5].
  5. A curved skateboard ramp rises from the ground; at a distance x meters from its foot, measured along the ground, its height is \displaystyle r(x) = \frac{1}{8}x^2 meters, for 0 \le x \le 4. Find the average rate of change of the height on [0, 2] and on [2, 4], and say which half of the ramp is steeper on average.

Comparing two functions

  1. Which of f(x) = x^2 and g(x) = 2^x has the greater average rate of change on [0, 4]? Which on [4, 6]?
  2. Ottoline’s balance after t years is H(t) = 1000 + 60t dollars and Barnaby’s is I(t) = 1000(1.05)^t dollars. Compare their average rates of change on [0, 5] and on [0, 10], to the nearest cent.

A symbolic interval

  1. For s(x) = x^2 + 2, find the average rate of change on [x, x + 1] as an expression in x, and use it to find the average rate on [3, 4].
  2. For the ramp \displaystyle r(x) = \frac{1}{8}x^2 of exercise 14, find the average rate of change on [x, x + 2] as an expression in x. Check it against both answers to exercise 14.

Explain

  1. Grahame works out the average rate of change of the ball’s height h(t) = -5t^2 + 25t + 2 on [0, 5], gets 0, and concludes that the ball did not move. Explain what the 0 does and does not say, using the one-second rates from Examples 2 and 6.
  2. Show that for f(x) = x^2 the average rate of change on [a, b], with a \ne b, is a + b. Use this to say which intervals give an average rate of 0, and check it against exercises 3, 4 and 5.
Answers
  1. 2. f(1) = 13 and f(6) = 23, so \displaystyle \frac{23 - 13}{6 - 1} = \frac{10}{5} = 2, the slope.
  2. \displaystyle -\frac{1}{3}. g(-3) = 9 and g(6) = 6, so \displaystyle \frac{6 - 9}{6 - (-3)} = \frac{-3}{9} = -\frac{1}{3}, the slope.
  3. 5. s(1) = 3 and s(4) = 18, so \displaystyle \frac{18 - 3}{3} = 5.
  4. -5. s(-4) = 18 and s(-1) = 3, so \displaystyle \frac{3 - 18}{-1 - (-4)} = \frac{-15}{3} = -5.
  5. 0. s(-2) = s(2) = 6, so \displaystyle \frac{6 - 6}{4} = 0.
  6. 20. w(2) = 10 and w(5) = 70, so \displaystyle \frac{70 - 10}{5 - 2} = \frac{60}{3} = 20.
  7. \displaystyle \frac{3}{2}, 6 and 24, from f(0) = 1, f(2) = 4, f(4) = 16, f(6) = 64. Each is 4 times the one before: moving 2 to the right multiplies every output of 2^x by 2^2 = 4, and so multiplies the change too.
  8. \displaystyle \frac{4}{3}. \displaystyle v(-1) = \frac{1}{3} and v(1) = 3, so \displaystyle \frac{3 - \frac{1}{3}}{2} = \frac{4}{3}.
  9. \displaystyle -\frac{56}{3} on [0, 3] and \displaystyle -\frac{7}{3} on [3, 6]. d(0) = 64, d(3) = 8, d(6) = 1, so \displaystyle \frac{8 - 64}{3} and \displaystyle \frac{1 - 8}{3}. The fall slows, since halving takes away half of an ever smaller amount.
  10. 15 meters per second on [0.5, 1.5] and -15 meters per second on [3.5, 4.5]. h(0.5) = 13.25, h(1.5) = 28.25, h(3.5) = 28.25, h(4.5) = 13.25. The intervals sit symmetrically about the peak at t = 2.5, so the rates are opposite.
  11. 10 meters per second on [0, 1] and -5 meters per second on [1, 3]. k(0) = 2, k(1) = 12, k(3) = 2, so \displaystyle \frac{12 - 2}{1} = 10 and \displaystyle \frac{2 - 12}{2} = -5. The ball is higher at 1 second than at launch, and lower at 3 seconds than at 1: it has passed its top.
  12. About 33.82 dollars per year on [0, 5] and 45.26 dollars per year on [5, 10]. B(5) \approx 669.11 and B(10) \approx 895.42, so \displaystyle \frac{669.11 - 500}{5} \approx 33.82 and \displaystyle \frac{895.42 - 669.11}{5} \approx 45.26.
  13. 562.5 users per month on [0, 4] and 2400 users per month on [4, 5]. N(0) = 150, N(4) = 2400, N(5) = 4800, so \displaystyle \frac{2250}{4} = 562.5 and \displaystyle \frac{4800 - 2400}{1} = 2400.
  14. \displaystyle \frac{1}{4} on [0, 2] and \displaystyle \frac{3}{4} on [2, 4], in meters of height per meter along the ground. r(0) = 0, \displaystyle r(2) = \frac{1}{2}, r(4) = 2, so \displaystyle \frac{1/2}{2} = \frac{1}{4} and \displaystyle \frac{2 - 1/2}{2} = \frac{3}{4}. The upper half is three times as steep on average.
  15. On [0, 4], f: \displaystyle \frac{16 - 0}{4} = 4 against \displaystyle \frac{16 - 1}{4} = \frac{15}{4} for g. On [4, 6], g: \displaystyle \frac{64 - 16}{2} = 24 against \displaystyle \frac{36 - 16}{2} = 10 for f.
  16. On [0, 5] Ottoline’s is greater, 60 against about 55.26 dollars per year; on [0, 10] Barnaby’s is greater, about 62.89 against 60. Ottoline’s rate is the slope 60 on every interval. I(5) \approx 1276.28 and I(10) \approx 1628.89, so \displaystyle \frac{276.28}{5} \approx 55.26 and \displaystyle \frac{628.89}{10} \approx 62.89.
  17. 2x + 1, and 7 on [3, 4]. s(x + 1) = x^2 + 2x + 3, and subtracting s(x) = x^2 + 2 leaves 2x + 1, divided by the width 1. At x = 3 this is 7, matching s(4) - s(3) = 18 - 11.
  18. \displaystyle \frac{x + 1}{4}. \displaystyle r(x + 2) - r(x) = \frac{(x + 2)^2 - x^2}{8} = \frac{4x + 4}{8}, and dividing by the width 2 gives \displaystyle \frac{x + 1}{4}. At x = 0 it gives \displaystyle \frac{1}{4} and at x = 2 it gives \displaystyle \frac{3}{4}, the two rates of exercise 14.
  19. The 0 says only that h(5) = h(0): the ball ends at the height it started from, 2 meters. It says nothing about the path. The five one-second rates are 20, 10, 0, -10 and -20 meters per second: the ball rose 30 meters to 32 by t = 2, went a little higher, and fell back. The rises and falls cancel, since \displaystyle \frac{20 + 10 + 0 - 10 - 20}{5} = 0.
  20. The top factors as b^2 - a^2 = (b - a)(b + a). Dividing out b - a, allowed because a \ne b, gives a + b. So the rate is 0 exactly when b = -a, on the intervals [-c, c] centered on 0. Adding 2 to every output changes no difference, so s(x) = x^2 + 2 has the same rates: 1 + 4 = 5, -4 + (-1) = -5 and -2 + 2 = 0, matching exercises 3, 4 and 5.
unattributed · CC-BY-SA-4.0