The Mean Value Theorem — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

On some toll roads, a ticket stamped at the entrance and the exit gives your average speed. If it says 75 miles per hour, the police know more than that: at some instant your speedometer read exactly 75. The Mean Value Theorem is the reason.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The average rate of change of $f$ on $[a, b]$ is

$\frac{f(b) - f(a)}{b - a}$
$\frac{f(a) + f(b)}{2}$
$f'(b) - f'(a)$
$\frac{b - a}{f(b) - f(a)}$

Show the answer

It is the slope of the secant line through the endpoints: change in output over change in input.

Correct: $\frac{f(b) - f(a)}{b - a}$ :
$\frac{f(a) + f(b)}{2}$ :
$f'(b) - f'(a)$ :
$\frac{b - a}{f(b) - f(a)}$ :

2. If $f$ is differentiable at $x = a$ , then

$f$ is continuous at $x = a$
$f$ has a corner at $x = a$
$f'(a) = 0$
$f$ is not continuous at $x = a$

Show the answer

Differentiability implies continuity, not the other way round.

Correct: $f$ is continuous at $x = a$ :
$f$ has a corner at $x = a$ :
$f'(a) = 0$ :
$f$ is not continuous at $x = a$ :

3. The Intermediate Value Theorem needs which hypothesis on $[a, b]$ ?

$f$ is continuous on $[a, b]$
$f$ is differentiable on $(a, b)$
$f(a) = f(b)$
$f$ is increasing

Show the answer

Continuity on the closed interval is the hypothesis of the IVT.

Correct: $f$ is continuous on $[a, b]$ :
$f$ is differentiable on $(a, b)$ :
$f(a) = f(b)$ :
$f$ is increasing:

Part 4 · See it

See it first

The graph of f(x) = x cubed minus 6x squared plus 9x plus 1 from (0, 1) to (4, 5). A dashed secant line joins the endpoints with slope 1. Tangent lines at c1 ≈ 0.85 and c2 ≈ 3.15 are parallel to it: at both points the derivative equals the average rate of change, 1. — The secant line through the endpoints has slope 1. At $c_1$ and $c_2$ the tangent lines have the same slope: $f'(c) = 1$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ the graph is one unbroken, smooth curve from $(a, f(a))$ to $(b, f(b))$ , with a tangent line at each inside point
Slide a line parallel to the secant line across the curvethe last place it touches the curve is a point where the tangent line is parallel to the secant line
Parallel lines have equal slopessome $c$ in $(a, b)$ has $f'(c)$ $= \frac{f(b) - f(a)}{b - a}$
A hypothesis fails (a jump, an asymptote, a corner or cusp inside)the theorem promises nothing, and the conclusion may be false

Part 6 · Key ideas

Key ideas

Mean Value Theorem: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then $f'(c)$ $= \frac{f(b) - f(a)}{b - a}$ for at least one $c$ in $(a, b)$ .
In words: somewhere inside, the instantaneous rate of change equals the average rate of change.
" $f$ is differentiable" covers both hypotheses, because differentiable implies continuous. Say it.
The theorem says a $c$ exists. It does not say where, and there may be more than one.
A corner at an endpoint is fine: differentiability is needed only on the open interval.

Part 7 · Misconception

A common mistake

The wrong idea: The Mean Value Theorem works whenever the average rate of change can be computed.

What actually happens: It needs both hypotheses: continuous on $[a, b]$ and differentiable on $(a, b)$ . For $|x|$ on $[-1, 1]$ the average rate is $0$ , but the slope is $\pm 1$ everywhere it exists. A justification must state the hypotheses before the conclusion.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Water level in a reservoir

The water level

h(t)

of a reservoir, in feet, is a differentiable function of time

t

, in days. The table gives selected values.

Water level $h(t)$ , in feet, at selected times $t$ , in days
$t$ (days)	0	3	7	12
$h(t)$ (feet)	40	46	38	40

1. Using the table, which statement is guaranteed for some time $c$ with $0 < c < 12$ ?

$h(c) = 35$
$h'(c) = 0$
$h'(c) = 2$ for a $c$ between 3 and 7
$h(c) > 46$

Show the answer

$\frac{h(12) - h(0)}{12 - 0}$ $= \frac{40 - 40}{12}$ $= 0$ . Since $h$ is differentiable, it is continuous on $[0, 12]$ and differentiable on $(0, 12)$ , so by the Mean Value Theorem some $c$ in $(0, 12)$ has $h'(c) = 0$ .

$h(c) = 35$ : 35 is below every value in the table. The IVT only promises values between two known values, such as between 38 and 46.
Correct: $h'(c) = 0$ : $h(0) = h(12) = 40$ , so the average rate of change on $[0, 12]$ is $0$ , and the Mean Value Theorem applies because $h$ is differentiable.
$h'(c) = 2$ for a $c$ between 3 and 7: On $[3, 7]$ the average rate is $\frac{38 - 46}{7 - 3} = -2$ , not $2$ . The sign matters.
$h(c) > 46$ : The table shows no level above 46. A higher level is possible between the readings, but nothing guarantees it.

2. On which interval does the Mean Value Theorem guarantee a time $c$ with $h'(c) = -2$ ?

$3 < c < 7$
$0 < c < 3$
$7 < c < 12$
$0 < c < 7$

Show the answer

The Mean Value Theorem gives $h'(c)$ equal to the average rate of change over an interval. On $[3, 7]$ that rate is $\frac{38 - 46}{7 - 3} = -2$ feet per day, and $h$ is differentiable, so it is continuous there.

Correct: $3 < c < 7$ : $\frac{h(7) - h(3)}{7 - 3}$ $= \frac{38 - 46}{4}$ $= -2$ .
$0 < c < 3$ : $\frac{46 - 40}{3} = 2$ : the right size, the wrong sign.
$7 < c < 12$ : $\frac{40 - 38}{5} = \frac{2}{5}$ .
$0 < c < 7$ : $\frac{38 - 40}{7} = -\frac{2}{7}$ . The theorem promises the average rate over the interval you choose, and this one is not $-2$ .

3. The Mean Value Theorem guarantees a time $c$ in $(7, 12)$ at which $h'(c)$ equals what value? Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\frac{h(12) - h(7)}{12 - 7}$ $= \frac{40 - 38}{5}$ $= \frac{2}{5}$ foot per day. $h$ is differentiable, so it is continuous on $[7, 12]$ and differentiable on $(7, 12)$ , and the Mean Value Theorem gives $h'(c) = \frac{2}{5}$ foot per day for some $c$ there.

Answer: $\frac{2}{5}$ ≈ 0.400 feet per day

4. A student writes: " $h'(c) = 0$ for some $c$ in $(0, 12)$ by the Mean Value Theorem, because $h(0) = h(12)$ ." What does this justification still need to earn the point?

The value of $c$ , found from the table
A check that $h(t) > 0$ on the interval
That $h$ is continuous on $[0, 12]$ and differentiable on $(0, 12)$
A second pair of times at which $h$ takes equal values, to confirm the first pair

Show the answer

A Mean Value Theorem justification names the theorem and checks both hypotheses: continuous on the closed interval, differentiable on the open one. " $h$ is differentiable, so it is also continuous" does both.

The value of $c$ , found from the table: The theorem says a $c$ exists; it does not ask you to find it, and the table cannot give it.
A check that $h(t) > 0$ on the interval: The sign of $h$ has nothing to do with the theorem's hypotheses.
Correct: That $h$ is continuous on $[0, 12]$ and differentiable on $(0, 12)$ : The theorem's hypotheses must be stated and checked. Here they follow from " $h$ is differentiable", which the problem gives.
A second pair of times at which $h$ takes equal values, to confirm the first pair: One interval with equal endpoint values is enough.

5. For $f(x) = x^3$ on $[0, 3]$ , find the value of $c$ guaranteed by the Mean Value Theorem. Give an exact answer.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Average rate: $\frac{f(3) - f(0)}{3 - 0}$ $= \frac{27}{3}$ $= 9$ . $f'(c) = 3c^2 = 9$ gives $c = \pm\sqrt{3}$ ; only $c = \sqrt{3} \approx 1.732$ lies in $(0, 3)$ .

Answer: $\sqrt{3}$ ≈ 1.732

6. $f$ is differentiable everywhere, with $f(1) = 3$ and $f(5) = 3$ . Which must be true?

$f(3) = 3$
$f'(3) = 0$ , at the midpoint of the interval
$f(c) = 0$ for some $c$ in $(1, 5)$
$f'(c) = 0$ for some $c$ in $(1, 5)$

Show the answer

$f$ is differentiable, so it is continuous on $[1, 5]$ and differentiable on $(1, 5)$ . The average rate of change is $0$ , so by the Mean Value Theorem $f'(c) = 0$ for some $c$ in $(1, 5)$ .

$f(3) = 3$ : Equal values at the ends say nothing about the value in the middle.
$f'(3) = 0$ , at the midpoint of the interval: A flat tangent line is guaranteed somewhere in $(1, 5)$ , not at the midpoint.
$f(c) = 0$ for some $c$ in $(1, 5)$ : No value below 3 is known, so the IVT gives no zero.
Correct: $f'(c) = 0$ for some $c$ in $(1, 5)$ : The average rate on $[1, 5]$ is $\frac{3 - 3}{4} = 0$ , and the Mean Value Theorem applies.

Part 9 · Summary

Summary

If

f

is continuous on

[a, b]

and differentiable on

(a, b)

, some

c

inside has

f'(c)

equal to the average rate of change: a tangent line parallel to the secant line. Check the hypotheses every time ("

f

is differentiable, so it is continuous") and name the theorem.

Part 10 · Up next

What comes next

Part 11 · Connections