Average Rate of Change | Math Coach Amy

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Key Ideas Easy Medium Hard

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

secant slope on [a, b]

What to Remember

Average rate of change

On the interval from a to b,

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

This is the slope of the secant line through the two points.

Units stay in the answer

If s is position in meters and t is time in seconds, the average rate is an average velocity in meters per second.

A line uses its slope

On a linear function, the average rate on every interval is the slope. On a curve, different intervals give different average rates.

The h form

\(\dfrac{f(a + h) - f(a)}{h}\) is the average rate from a to \(a + h\). The limit as \(h \to 0\) is the instantaneous rate.

Level 1

Easy

Two points on a function you can evaluate.

Example

Find the average rate of change of \(f(x) = x^2\) on \([1, 4]\).

\(\dfrac{f(4) - f(1)}{4 - 1}\)

\(f(4) = 16\)

\(f(1) = 1\)

\(\dfrac{16 - 1}{3} = \dfrac{15}{3}\)

\(5\)

Answer: \(5\)

Find the average rate of change of \(f(x) = x^2\) on \([1, 3]\).

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\(\dfrac{f(3) - f(1)}{3 - 1}\)

\(f(3) = 9\)

\(f(1) = 1\)

\(\dfrac{8}{2} = 4\)

Answer: \(4\)

Find the average rate of change of \(f(x) = 3x + 1\) on \([2, 6]\).

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\(\dfrac{f(6) - f(2)}{6 - 2}\)

\(f(6) = 19\)

\(f(2) = 7\)

\(\dfrac{12}{4} = 3\)

The function is linear with slope 3, so every interval has average rate 3.

Answer: \(3\)

Find the average rate of change of \(f(x) = x^3\) on \([0, 2]\).

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\(\dfrac{f(2) - f(0)}{2 - 0}\)

\(f(2) = 8\)

\(f(0) = 0\)

\(\dfrac{8}{2} = 4\)

Answer: \(4\)

A particle’s position is \(s(t) = t^2\) meters at time t seconds. Find the average velocity on \([1, 4]\).

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\(\dfrac{s(4) - s(1)}{4 - 1}\)

\(s(4) = 16\)

\(s(1) = 1\)

\(\dfrac{15}{3} = 5\)

5 meters per second

Answer: 5 meters per second

Find the average rate of change of \(f(x) = \sqrt{x}\) on \([4, 9]\).

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\(\dfrac{f(9) - f(4)}{9 - 4}\)

\(f(9) = 3\)

\(f(4) = 2\)

\(\dfrac{1}{5}\)

Answer: \(\dfrac{1}{5}\)

Level 2

Medium

A table, a sign, or a comparison with nearby rates.

Example

The table gives the position s(t), in meters, of a particle at time t seconds. Find the average velocity on \([2, 8]\).

\(t\)	0	2	5	8
\(s(t)\)	3	11	20	26

\(\dfrac{s(8) - s(2)}{8 - 2}\)

\(s(8) = 26\)

\(s(2) = 11\)

\(\dfrac{15}{6} = \dfrac{5}{2}\)

\(\dfrac{5}{2}\) meters per second

Answer: \(\dfrac{5}{2}\) meters per second

Use the same position table. Find the average velocity on \([0, 5]\).

\(t\)	0	2	5	8
\(s(t)\)	3	11	20	26

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\(\dfrac{s(5) - s(0)}{5 - 0}\)

\(s(5) = 20\)

\(s(0) = 3\)

\(\dfrac{17}{5}\)

\(\dfrac{17}{5}\) meters per second

Answer: \(\dfrac{17}{5}\) meters per second

Find the average rate of change of \(f(x) = x^2 + 1\) on \([-1, 3]\).

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\(\dfrac{f(3) - f(-1)}{3 - (-1)}\)

\(f(3) = 10\)

\(f(-1) = 2\)

\(\dfrac{8}{4} = 2\)

Answer: \(2\)

The average rate of change of \(f(x) = x^2\) on \([1, 3]\) is

A. \(2\)

B. \(4\)

C. \(6\)

D. \(8\)

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\(\dfrac{f(3) - f(1)}{3 - 1}\)

\(\dfrac{9 - 1}{2} = 4\)

Answer: B. 4

The position of a particle is \(s(t) = t^2 - 4t\) meters. Find the average velocity from \(t = 1\) to \(t = 5\).

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\(\dfrac{s(5) - s(1)}{5 - 1}\)

\(s(5) = 25 - 20 = 5\)

\(s(1) = 1 - 4 = -3\)

\(\dfrac{5 - (-3)}{4} = \dfrac{8}{4}\)

\(2\)

2 meters per second

Answer: 2 meters per second

Find the average rate of change of \(f(x) = \dfrac{1}{x}\) on \([1, 4]\).

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\(\dfrac{f(4) - f(1)}{4 - 1}\)

\(f(4) = \dfrac{1}{4}\)

\(f(1) = 1\)

\(\dfrac{\tfrac{1}{4} - 1}{3} = \dfrac{-\tfrac{3}{4}}{3}\)

\(-\dfrac{3}{4} \cdot \dfrac{1}{3} = -\dfrac{1}{4}\)

Answer: \(-\dfrac{1}{4}\)

Level 3

Hard

Compare intervals, solve for a missing value, or let the interval shrink.

Example

The table gives values of g. On which interval is the average rate of change greater, \([1, 4]\) or \([4, 6]\)?

\(x\)	1	4	6
\(g(x)\)	2	14	18

Average rate on \([1, 4]\):

\(\dfrac{g(4) - g(1)}{4 - 1} = \dfrac{14 - 2}{3} = 4\)

Average rate on \([4, 6]\):

\(\dfrac{g(6) - g(4)}{6 - 4} = \dfrac{18 - 14}{2} = 2\)

\(4 > 2\)

Answer: \([1, 4]\)

For \(f(x) = x^2\), write the average rate of change on \([2, 2 + h]\) with \(h \neq 0\), and then find the limit as \(h \to 0\).

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\(\dfrac{f(2 + h) - f(2)}{h}\)

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

\(f(2) = 4\)

\(\dfrac{4h + h^2}{h} = 4 + h\) for \(h \neq 0\)

The limit as \(h \to 0\) is

\(4\)

Answer: \(4 + h\), then \(4\)

The average rate of change of f on \([1, 5]\) is 3, and \(f(1) = 4\). Find \(f(5)\).

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\(\dfrac{f(5) - f(1)}{5 - 1} = 3\)

\(\dfrac{f(5) - 4}{4} = 3\)

\(f(5) - 4 = 12\)

\(f(5) = 16\)

Answer: \(16\)

A particle has \(s(1) = 2\) meters. The average velocity on \([1, 3]\) is 6 meters per second. Find \(s(3)\).

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\(\dfrac{s(3) - s(1)}{3 - 1} = 6\)

\(\dfrac{s(3) - 2}{2} = 6\)

\(s(3) - 2 = 12\)

\(s(3) = 14\)

Answer: \(14\) meters

The temperature of a liquid is 18°C at \(t = 1\) hour and 10°C at \(t = 5\) hours. Find the average rate of change of temperature, and say what the sign means.

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\(\dfrac{10 - 18}{5 - 1}\)

\(\dfrac{-8}{4} = -2\)

The temperature decreased by 2°C per hour, on average.

Answer: \(-2\) °C per hour

Find the average rate of change of \(f(x) = x^3 - x\) on \([0, 2]\).

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\(\dfrac{f(2) - f(0)}{2 - 0}\)

\(f(2) = 8 - 2 = 6\)

\(f(0) = 0\)

\(\dfrac{6}{2} = 3\)

Answer: \(3\)

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