Unit 1, lesson 8

Average Rate of Change and Secant Lines

How the average rate of change measures output change per unit of input change over an interval, why it equals the slope of a secant line, and how to read its sign and units.

  • Theory part 15
  • Practice part 16
  • Rates and Graph Behavior

A drive of 240 kilometers that takes 3 hours averages 80 kilometers per hour, even with a stop for lunch and a fast stretch on the highway. The average looks only at the start and the end and ignores everything in between. For any function the same comparison is the slope of a line through two points of its graph. Shrinking the gap between those two points is how calculus defines speed at a single instant.

Key terms

Average rate of change

Definition Let ff be defined at aa and at bb, with a≠ba \ne b. The average rate of change of ff from aa to bb is

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

When a<ba < b it is also called the average rate of change of ff on the interval [a,b][a, b]. Only the values at aa and bb enter the formula, not the values in between.

In plain words It is how much the output changed per unit of input, on average, over the stretch from aa to bb.

Example For f(x)=x2f(x) = x^2 on [1,4][1, 4] the average rate of change is

f(4)−f(1)4−1=16−13=5.\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = 5.

On average the output went up by 5 for every unit of input.

Output change over nonzero input change

Definition Write Δx=b−a\Delta x = b - a for the change in input and Δy=f(b)−f(a)\Delta y = f(b) - f(a) for the change in output. The average rate of change is the quotient

ΔyΔx=f(b)−f(a)b−a,\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a},

which is defined only when Δx≠0\Delta x \ne 0. Both differences must be taken in the same order. Swapping aa and bb in both of them changes nothing, since the numerator and the denominator both change sign. Writing b=a+hb = a + h with h≠0h \ne 0 gives the difference quotient

f(a+h)−f(a)h,\frac{f(a + h) - f(a)}{h},

the form calculus uses for velocity and tangent lines.

In plain words Subtract the outputs, subtract the inputs in the same order, and divide. The input change can never be zero, because there is no rate over a stretch of length zero.

Example For f(x)=x2f(x) = x^2 with a=5a = 5 and b=2b = 2, keeping the order consistent gives

f(2)−f(5)2−5=4−25−3=7,\frac{f(2) - f(5)}{2 - 5} = \frac{4 - 25}{-3} = 7,

the same as 25−45−2=7\frac{25 - 4}{5 - 2} = 7. Mixing the orders, as in 25−42−5\frac{25 - 4}{2 - 5}, gives the wrong sign, −7-7. With a=b=5a = b = 5 the quotient would be 00\frac{0}{0}, which is undefined.

Secant slope

Definition For a≠ba \ne b in the domain of ff, the secant line of the graph of ff from aa to bb is the line through the two graph points (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)). Its slope is the average rate of change of ff from aa to bb, and its equation is

y=f(a)+f(b)−f(a)b−a (x−a).y = f(a) + \frac{f(b) - f(a)}{b - a}\,(x - a).

In plain words Draw the straight line through two points of the graph. Its slope is the average rate, and it is the graph of the linear function that agrees with ff at aa and at bb.

Example For f(x)=xf(x) = \sqrt{x} with a=1a = 1 and b=9b = 9 the points are (1,1)(1, 1) and (9,3)(9, 3), so the secant slope is

3−19−1=14.\frac{3 - 1}{9 - 1} = \frac{1}{4}.

The secant line is y=1+14(x−1)y = 1 + \frac{1}{4}(x - 1), or y=14x+34y = \frac{1}{4}x + \frac{3}{4}. Between the two points the curve runs above this line. At x=4x = 4, for instance, the curve has height 4=2\sqrt{4} = 2 and the line has height 1.751.75.

Units of a rate

Definition If the input is measured in units UU and the output in units VV, the average rate of change is measured in VV per UU, because it divides a change in output by a change in input. When the input is time and the output is position, the average rate of change is the average velocity.

In plain words Carry the units through the division. The result reads as so many output units for each input unit.

Example The temperature outside was 12 degrees Celsius at 7 in the morning and 21 degrees Celsius at 10 in the morning. Its average rate of change over those 3 hours is

21−1210−7=3 degrees Celsius per hour.\begin{aligned} &\frac{21 - 12}{10 - 7} \\[0.8em] &\quad = 3 \text{ degrees Celsius per hour.} \end{aligned}

Signed rate

Definition For a<ba < b the average rate of change is positive when f(b)>f(a)f(b) > f(a), negative when f(b)<f(a)f(b) < f(a), and zero when f(b)=f(a)f(b) = f(a). A positive rate means the secant line rises from left to right, and a negative rate means it falls. The sign describes only the net change from aa to bb. In between, ff may rise and fall, so a rate of 00 does not mean that ff is constant.

In plain words A positive rate means a net gain and a negative rate a net loss. A rate of zero means you ended where you started, not that nothing happened.

Example A ball thrown straight up is at height h(t)=20t−5t2h(t) = 20t - 5t^2 meters after tt seconds.

IntervalAverage rate of change
[0,1][0, 1]15−01=15\frac{15 - 0}{1} = 15 meters per second
[1,3][1, 3]15−152=0\frac{15 - 15}{2} = 0 meters per second
[2,4][2, 4]0−202=−10\frac{0 - 20}{2} = -10 meters per second

On [1,3][1, 3] the rate is 00 even though the ball rose to 20 meters at t=2t = 2 and came back down.

Comparing intervals

Definition To compare how fast ff changes on two intervals, compare its average rates of change on them. Comparing the output changes Δy\Delta y alone is fair only when the intervals have the same length. A linear function has the same average rate of change on every interval, equal to its slope. For a function that is not linear the average rate is not the same on every interval, and the way it varies shows where the graph gets steeper or flatter.

In plain words Divide before you compare. A larger change over a longer stretch can still be a slower rate.

Example For f(x)=x2f(x) = x^2 the intervals [0,1][0, 1], [1,2][1, 2] and [2,3][2, 3] have average rates 11, 33 and 55, so the graph gets steeper as xx grows. The interval [0,3][0, 3] has the largest output change, 99, yet its average rate is only 93=3\frac{9}{3} = 3, smaller than the rate 55 on [2,3][2, 3].

Check yourself

  1. A hiker is at an elevation of 1250 meters after 2 hours of walking and 1730 meters after 5 hours. Find the average rate of change of the elevation, with units.
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The average rate of change is

1730−12505−2=4803=160 meters per hour.\begin{aligned} &\frac{1730 - 1250}{5 - 2} = \frac{480}{3} \\[0.8em] &\quad = 160 \text{ meters per hour.} \end{aligned}
  1. For f(x)=x3−4xf(x) = x^3 - 4x, find the average rate of change on [−2,2][-2, 2]. Does the answer mean that ff is constant on [−2,2][-2, 2]?
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Both endpoint values are zero, f(−2)=0f(-2) = 0 and f(2)=0f(2) = 0, so the average rate of change is 0−04=0\frac{0 - 0}{4} = 0. The function is not constant, since f(−1)=3f(-1) = 3 and f(1)=−3f(1) = -3. The values 00, 33, −3-3, 00 at x=−2,−1,1,2x = -2, -1, 1, 2 show the graph rising, falling and rising again, yet it ends at the same height where it started.

  1. For f(x)=x2+3xf(x) = x^2 + 3x, simplify the average rate of change from 22 to 2+h2 + h, where h≠0h \ne 0. What number does it approach as hh gets close to 00?
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Since f(2)=10f(2) = 10, for h≠0h \ne 0

(2+h)2+3(2+h)−10h=7h+h2h=7+h.\begin{aligned} &\frac{(2 + h)^2 + 3(2 + h) - 10}{h} \\[0.8em] &\quad = \frac{7h + h^2}{h} = 7 + h. \end{aligned}

As hh gets close to 00 this approaches 77, which in calculus is the slope of the tangent line at (2,10)(2, 10).

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