Unit 5 · Topic 5.1 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 5.1: Using the Mean Value Theorem

Q: Does the theorem tell me where c is?

No. It only says at least one c exists in (a, b). When you have a formula for f, you can find c by solving f'(c) equal to the average rate; with only a table, you cannot.

Q: Why does it need differentiable only on the open interval?

The tangent line it promises sits strictly inside the interval, so the derivative only has to exist there. x on [0, 4] has no derivative at 0 and still meets the hypotheses.

Q: Can I use the theorem on a table of values?

Yes, if the problem says the function is differentiable (or continuous and differentiable) on the interval. Pick two rows, compute the average rate between them, and the theorem gives a point between those two inputs where the derivative equals it.

Q: How is this different from the Intermediate Value Theorem?

The Intermediate Value Theorem needs only continuity and is about values: f(c) hits every number between f(a) and f(b). The Mean Value Theorem also needs differentiability and is about the derivative: f'(c) equals the average rate.

Q: What if the solution for c lands on an endpoint?

Then it is not a value the theorem promises, because the theorem's c is strictly between a and b. Keep only the solutions inside (a, b).

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 5: 15 to 20% of the multiple-choice section

What you'll be able to do

Justify a conclusion about a function by applying the Mean Value Theorem over an interval, after checking that its hypotheses hold.

Key vocabulary

Term	What it means	The exam trap
Mean Value Theorem	If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , there is a point in $(a, b)$ where the instantaneous rate of change equals the average rate of change over $[a, b]$ .	The conclusion is about a derivative, a rate, not about a value of the function.
Continuous on $[a, b]$	The function is continuous at every point of the closed interval, endpoints included.	Look for any input in the interval where the function has no value or jumps, not only at the endpoints.
Differentiable on $(a, b)$	The function has a derivative at every point strictly between a and b.	A graph can be unbroken and still fail here, so check for sharp points as well as breaks.
Average rate of change	The change in a function's output over an interval divided by the change in its input, $\frac{f (b) - f (a)}{b - a}$ .	Divide by the change in the input; the change in the output alone is not a rate.
Instantaneous rate of change	The rate at one input, the derivative $f^{'} (c)$ .	Read whether a question is about one input or a whole interval before you pick a formula.

Graph of f(x) = x cubed minus 6x squared plus 9x plus 1 for x from 0 to 4. A secant line joins (0, 1) and (4, 5) and has slope 1. Dashed tangent lines at x = 0.85 and x = 3.15, both inside the interval, also have slope 1 and run parallel to the secant.

Part 1 Think first Two cameras, one ticket?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A car passes a highway camera at mile marker 12 at 2:00 p.m. and a second camera at mile marker 42 at 2:24 p.m. The speed limit is 65 miles per hour, and nobody saw the car's speedometer. Must the car have been going exactly 75 miles per hour at some instant between the two cameras?

In short. The Mean Value Theorem says that 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 $c$ with $a < c < b$ .

Check both hypotheses first: continuous on the closed interval $[a, b]$ , differentiable on the open interval $(a, b)$ .
The conclusion is about the derivative: the instantaneous rate at some $c$ equals the average rate over the whole interval.
The theorem says such a $c$ exists, strictly between $a$ and $b$ ; it does not say where, or that there is only one.

Part 2 Learn it Three short ideas, each checked as you go

5a. What the theorem promises

The Mean Value Theorem joins the two rates from Unit 2. Suppose $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ . Then there is at least one number $c$ with $a < c < b$ at which

f^{'} (c) = \frac{f (b) - f (a)}{b - a} .

The right side is the average rate of change of $f$ over $[a, b]$ . The left side is the instantaneous rate of change at $c$ . So somewhere strictly inside the interval, the rate at an instant equals the average rate over the whole interval.

On a graph, the average rate is the slope of the secant line through the endpoints, and $f^{'} (c)$ is the slope of the tangent line at $c$ . The theorem says some tangent line inside the interval is parallel to that secant line. The cubic at the top of the page has two of them.

Read carefully what it does not say. It does not say where $c$ is; the midpoint is not special. It does not say there is only one $c$ . And it is a statement about $f^{'} (c)$ , a rate. A statement about a value $f (c)$ lying between $f (a)$ and $f (b)$ belongs to a different theorem, the Intermediate Value Theorem from Unit 1.

Quick check · Multiple choice

No calculator. A function $f$ is continuous on $[2, 6]$ and differentiable on $(2, 6)$ , with $f (2) = 7$ and $f (6) = 19$ . Which statement does the Mean Value Theorem guarantee?

5b. Check the hypotheses before you use it

The theorem promises nothing unless both hypotheses hold, so check them before you use it.

Continuous on $[a, b]$ , endpoints included. A jump, a hole or a vertical asymptote anywhere in $[a, b]$ breaks it.
Differentiable on $(a, b)$ , the open interval. A corner, a cusp or a vertical tangent anywhere strictly between $a$ and $b$ breaks it. Nothing is asked about the derivative at $a$ or $b$ themselves.

Polynomials, $\sin x$ , $\cos x$ and $e^{x}$ are differentiable at every real number, so they pass on any interval. A rational function passes on any interval that contains no input where its denominator is 0.

When a hypothesis fails, the conclusion can fail with it. Take $f (x) = | x |$ on $[- 2, 1]$ . The average rate is $\frac{1 - 2}{1 - (- 2)} = - \frac{1}{3}$ , but $f^{'} (x)$ is only ever $- 1$ or $1$ , so no $c$ works. The corner at 0 is the reason. (Sometimes the conclusion happens to hold anyway; the theorem just no longer promises it.)

Recall from Unit 2 that a function that is differentiable at a point is continuous there. So if you are told $f$ is differentiable on all of $[a, b]$ , both hypotheses hold at once.

Quick check · Sort it

No calculator. Tap a function on an interval, then tap whether the Mean Value Theorem applies to it there.

Both hypotheses hold

A hypothesis fails

5c. Writing the justification

When a question says justify, a complete answer using this theorem has three parts. State the hypotheses and why they hold. Work out the average rate of change. Then state the conclusion about the function by name, on the open interval, with units if there is a context.

Suppose $V (t)$ is the volume of water in a tank, in gallons, $t$ minutes after noon, and $V$ is differentiable for $0 \leq t \leq 10$ . You know only that $V (2) = 50$ and $V (6) = 38$ . A full justification reads:

$V$ is differentiable on $[2, 6]$ , so it is continuous on $[2, 6]$ and differentiable on $(2, 6)$ . $\frac{V (6) - V (2)}{6 - 2} = \frac{38 - 50}{4} = - 3$ . By the Mean Value Theorem, there is a time $c$ with $2 < c < 6$ at which $V^{'} (c) = - 3$ gallons per minute.

That is how the theorem works on a table of values: the problem tells you the function is differentiable, and you choose two rows. Without that sentence, a table tells you nothing about what happens between its rows, and the theorem cannot be used.

Name the function every time. "It has a slope of $- 3$ " leaves a reader guessing whether you mean $V$ or $V^{'}$ , or the graph of either.

Quick check · Decide and justify

No calculator. A function $g$ is continuous on $[1, 7]$ , with $g (1) = 6$ and $g (7) = 6$ . Nothing else is known about $g$ . Must there be a number $c$ in $(1, 7)$ at which $g^{'} (c) = 0$ ?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Using the theorem without checking that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ .	A corner, a cusp or a break inside the interval can make the conclusion false, as $\| x \|$ on $[- 2, 1]$ shows.	State both hypotheses with a reason, such as " $f$ is a polynomial" or " $f$ is differentiable, so it is continuous", before you apply the theorem.
Taking $c$ to be the midpoint of the interval, or assuming only one $c$ works.	The theorem says at least one $c$ exists in $(a, b)$ . It gives no location, and there can be several.	Solve $f^{'} (c)$ equal to the average rate, and keep every solution that lies strictly between $a$ and $b$ .
Concluding that $f (c)$ equals some number, instead of $f^{'} (c)$ .	That mixes this theorem up with the Intermediate Value Theorem, which is about values of $f$ , not its rate of change.	Mean Value Theorem: a derivative equals an average rate. Intermediate Value Theorem: a function value lies between two others.

Part 3 Watch it done Finding the number the theorem promises

Watch each step, then do the last one yourself.

Let $f (x) = x^{3} + x$ . Show that the Mean Value Theorem applies to $f$ on $[- 1, 2]$ , and find the value of $c$ it guarantees.

Step 1 · Check the hypotheses
$f$ is a polynomial, so it is continuous on $[- 1, 2]$ and differentiable on $(- 1, 2)$ . The Mean Value Theorem applies.
Why this step?
Why is "polynomial" enough to settle both hypotheses?
A polynomial has a derivative at every real number, and a function that is differentiable at a point is also continuous there (Unit 2). So one fact covers both conditions on any interval.
Step 2 · Find the average rate of change
$f (2) = 8 + 2 = 10$ and $f (- 1) = - 1 - 1 = - 2$ , so $\frac{f (2) - f (- 1)}{2 - (- 1)} = \frac{10 - (- 2)}{3} = 4$ .
Step 3 · Set the derivative equal to it
$f^{'} (x) = 3 x^{2} + 1$ , so solve $3 c^{2} + 1 = 4$ . Then $3 c^{2} = 3$ , $c^{2} = 1$ , and $c = 1$ or $c = - 1$ .
Your step · Calculate

No calculator. Finish it. The equation $3 c^{2} + 1 = 4$ has solutions $c = 1$ and $c = - 1$ . Which value of $c$ does the Mean Value Theorem guarantee for $f (x) = x^{3} + x$ on $[- 1, 2]$ ?

Your answer

Part 4 Try it A new function and a cyclist

Try it · Calculate

No calculator. Let $f (x) = \sqrt{x}$ . Find the value of $c$ in $(0, 9)$ at which $f^{'} (c)$ equals the average rate of change of $f$ over $[0, 9]$ . Give an exact decimal.

Your answer

Try it · Multiple choice

No calculator. A cyclist's velocity $v (t)$ , in meters per second, is a differentiable function of time $t$ , in seconds. Her velocity is 2 meters per second at $t = 0$ and 8 meters per second at $t = 10$ . Which statement must be true?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. A pot of soup sits on a stove, and the cook adjusts the heat as it cooks. The soup's temperature, in degrees Celsius, is given by a differentiable function $T$ of time $t$ , in minutes, for $0 \leq t \leq 20$ . Selected values: $T (0) = 20$ , $T (4) = 44$ , $T (10) = 62$ , $T (15) = 57$ and $T (20) = 90$ .

(a) Explain why there must be a time $t$ , with $4 < t < 10$ , at which $T^{'} (t) = 3$ .

(b) Must there be a time $t$ , with $0 < t < 20$ , at which the soup is cooling? Justify your answer.

(c) The cook says the soup's temperature never rose faster than 5 degrees Celsius per minute. Use the Mean Value Theorem to explain why the cook's statement is false.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $T$ is differentiable, so it is continuous on $[4, 10]$ and differentiable on $(4, 10)$ . $\frac{T (10) - T (4)}{10 - 4} = \frac{62 - 44}{6} = 3$ . By the Mean Value Theorem, there is a time $t$ with $4 < t < 10$ at which $T^{'} (t) = 3$ degrees Celsius per minute.

(b) Yes. $T$ is continuous on $[10, 15]$ and differentiable on $(10, 15)$ , and $\frac{T (15) - T (10)}{15 - 10} = \frac{57 - 62}{5} = - 1$ . By the Mean Value Theorem, $T^{'} (t) = - 1$ for some $t$ with $10 < t < 15$ . At that time the temperature is decreasing at 1 degree Celsius per minute, so the soup is cooling.

(c) $T$ meets the hypotheses on $[0, 4]$ , and $\frac{T (4) - T (0)}{4 - 0} = \frac{44 - 20}{4} = 6$ . By the Mean Value Theorem, $T^{'} (t) = 6$ for some $t$ with $0 < t < 4$ . At that time the temperature was rising at 6 degrees Celsius per minute, which is faster than 5, so the cook's statement is false.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

(a) States that $T$ is continuous on $[4, 10]$ and differentiable on $(4, 10)$ , with the reason that $T$ is differentiable.
(a) Computes $\frac{T (10) - T (4)}{10 - 4} = 3$ and concludes, by the Mean Value Theorem, that $T^{'} (t) = 3$ for some $t$ in $(4, 10)$ .
(b) Answers yes, with the hypotheses stated, using $[10, 15]$ and its average rate of $- 1$ to conclude $T^{'} (t) = - 1 < 0$ for some $t$ in $(10, 15)$ .
(c) Computes an average rate above 5 on an interval from the table (6 on $[0, 4]$ , or 6.6 on $[15, 20]$ ).
(c) States that $T$ meets the hypotheses there and concludes $T^{'} (t)$ equals that rate for some $t$ inside the interval, in degrees Celsius per minute.

Part 6 Check yourself A drone over a hill

The two case questions use the same table. Answer them from it.

A delivery drone climbs over a hill and then descends toward a landing pad. Its altitude above the pad is $H (t)$ feet, $t$ seconds after a timer starts, and $H$ is differentiable for $0 \leq t \leq 12$ . Some values are in the table.

Altitude of the drone above the landing pad
$t$ (seconds)	$H (t)$ (feet)
0	20
2	26
5	23
9	11
12	10

Case question 1 · Calculate

No calculator. Applying the Mean Value Theorem to the interval $2 \leq t \leq 12$ guarantees a time $c$ , with $2 < c < 12$ , at which $H^{'} (c)$ equals what value, in feet per second?

Your answer

Case question 2 · Multiple choice

No calculator. Applying the Mean Value Theorem to which interval guarantees a time at which the drone's altitude is changing at a rate of $- 3$ feet per second?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student applies the Mean Value Theorem to $h (x) = \frac{1}{x - 2}$ on $[1, 4]$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. For each function on its interval, find every $c$ in the open interval where $f^{'} (c)$ equals the average rate of change. Tap a function, then its result.

Part 7 Keep it Bring back earlier ideas

The theorem leans on the meaning of a derivative in context and on its older cousin from Unit 1. Pulling them back up is what makes them last.

Bring it back · Fill the statement

No calculator. Tap a blank, then tap an answer.

A tank holds $W (t)$ liters of water $t$ minutes after it starts to drain. The unit of $W^{'} (t)$ is , and $W^{'} (3) = - 2$ means that at $t = 3$ the amount of water in the tank is .

Bring it back · Decide and justify

No calculator. A function $f$ is continuous on $[0, 4]$ , with $f (0) = - 3$ and $f (4) = 5$ . Must there be a number $c$ in $(0, 4)$ at which $f (c) = 0$ ?

Which reason supports your answer?

In three lines

The Mean Value Theorem needs $f$ continuous on $[a, b]$ and differentiable on $(a, b)$ ; check both before using it.
It then promises at least one $c$ in $(a, b)$ where $f^{'} (c)$ equals the average rate $\frac{f (b) - f (a)}{b - a}$ .
A justification states the hypotheses, the average rate and the conclusion about the named function, with units.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question says "justify" and the Mean Value Theorem fits, write three things: the hypotheses with a reason (" $f$ is differentiable, so it is continuous on $[a, b]$ and differentiable on $(a, b)$ "), the average rate with its arithmetic shown, and the conclusion naming the function, the open interval and the units. Saying "by the MVT" without the hypotheses leaves the argument unfinished.

Questions students ask

Does the theorem tell me where c is?

No. It only says at least one $c$ exists in $(a, b)$ . When you have a formula for $f$ , you can find $c$ by solving $f^{'} (c)$ equal to the average rate; with only a table, you cannot.

Why does it need differentiable only on the open interval?

The tangent line it promises sits strictly inside the interval, so the derivative only has to exist there. $\sqrt{x}$ on $[0, 4]$ has no derivative at 0 and still meets the hypotheses.

Can I use the theorem on a table of values?

Yes, if the problem says the function is differentiable (or continuous and differentiable) on the interval. Pick two rows, compute the average rate between them, and the theorem gives a point between those two inputs where the derivative equals it.

How is this different from the Intermediate Value Theorem?

The Intermediate Value Theorem needs only continuity and is about values: $f (c)$ hits every number between $f (a)$ and $f (b)$ . The Mean Value Theorem also needs differentiability and is about the derivative: $f^{'} (c)$ equals the average rate.

What if the solution for c lands on an endpoint?

Then it is not a value the theorem promises, because the theorem's $c$ is strictly between $a$ and $b$ . Keep only the solutions inside $(a, b)$ .

Course alignment, for teachers

AP Calculus AB topic 5.1, Unit 5: Analytical Applications of Differentiation.

Skill 3.E: Provide reasons or rationales for solutions and conclusions.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.