Unit 5The Mean Value Theorem (MVT)

Concept 4.125Built for May 2027 AP Exam

The Mean Value Theorem (MVT)

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The Mean Value Theorem says that a smooth curve must copy its own average slope somewhere inside the interval. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then some $c$ in $(a,b)$ has $f'(c) = \frac{f(b) - f(a)}{b - a}$ . The right side is a number you compute from the endpoints; the left side is an instantaneous slope that the theorem promises exists. Almost every exam version is a justification: name continuity, name differentiability, compute the average rate, then conclude. The conclusion is existence, so the answer is a sentence rather than a value, and the sentence is what gets scored.

On the AP Exam

Five of the nine released exams since 2017 contain a part that requires citing this theorem, and the wording is nearly identical each time.

What you’ll see: A table of values, a graph, or a differentiable function with two known points, followed by "must there exist a value $c$ ... such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ " or "explain why there must be at least one time at which the rate equals the average rate".
What you’ll do: Compute the average rate of change from the two endpoints, state the two hypotheses, and conclude that some interior point has that instantaneous rate.
Memory hook: Average somewhere becomes instant somewhere. The chord is copied by a tangent.
Use it on the exam: For MCQs: Solve $f'(c) = \frac{f(b)-f(a)}{b-a}$ directly and keep only the roots inside the open interval. Distractors come from roots outside it. For FRQs: The words "there must exist" are the instruction to justify rather than to solve. Write continuity, differentiability, and the average rate before the conclusion sentence.

Concept Review

The Secant Gets Copied Somewhere

Figure 1 — The secant across the interval has a slope; the theorem promises a tangent inside with the same slope. Rolle's theorem is the case where that slope is zero.

Mean Value Theorem: a tangent parallel to the secant exists inside the interval

The Statement and Its Two Halves

$f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \text{ in } (a,b)$

The right-hand side depends only on the two endpoints, so it is a single number you can compute before writing anything else. The left-hand side is a slope that may not be findable by formula, which is why the conclusion is existential. Splitting the sentence in two is the fastest way to avoid mixing the roles: compute the number, then say what has to equal it.

Rolle's Theorem Is the Level Case

When $f(a) = f(b)$ , the average rate is zero and the conclusion becomes $f'(c) = 0$ . That is Rolle's theorem, and the ordering of the two topics on this site follows the logic: the special case is easier to check because the endpoint comparison can be done by eye. Everything the previous topic teaches about verifying hypotheses applies here unchanged, with one extra computation added.

The Sentence the Rubric Wants

Three clauses, in this order, earn the points. Continuity on the closed interval. Differentiability on the open interval. The existence of $c$ with the two slopes equal. A student who writes only the third clause has answered the question but not justified it, and the released rubrics separate those two achievements.

What the stem asks	What the first line must be	What the last line must be
"must there exist $c$ with $f'(c) = k$ "	the average rate equals $k$	therefore such a $c$ exists
"explain why there must be a time"	$f$ is differentiable, hence continuous	by the MVT there is such a time
"does the MVT guarantee a value"	hypotheses hold or fail, with a reason	yes or no, tied to that reason
"find $c$ "	the average rate	solve the equation and keep roots inside

Common Misconception Check

Myth 1: "The theorem finds the location of $c$ ." Truth: It guarantees existence. Solving for $c$ is a separate request, and some functions make it impossible in closed form.
Myth 2: "Differentiability on the closed interval is what the theorem needs." Truth: Continuity is the closed-interval requirement and differentiability is the open-interval one. Swapping them is a written error the rubric notices.
Myth 3: "A function with a corner cannot satisfy the MVT." Truth: A corner can sit at an endpoint and the theorem still applies, because differentiability is only required inside the open interval.

Topic-Specific Study Method

Compute the Average, Then Claim the Instant

Borrowed idea: Between any two points on a smooth curve, some tangent is parallel to the chord. Is that the same as saying the average and the instantaneous rate agree somewhere? Yes, and that is the whole theorem. The chord is the average and the tangent is the instant. Rolle's theorem is the case where the chord is flat. Why does the exam ask for existence instead of a value? Because the location of $c$ is usually not knowable.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table with two rows of data	$t$ and $f(t)$ columns	average rate from the two endpoints	compute, state hypotheses, conclude
"acceleration equals average acceleration"	velocity described in the stem	MVT applied to $v$	average rate of $v$ , then the claim
"average velocity" then "instantaneous"	two vocabulary words	the same theorem in motion language	one computation, one sentence
Graph of $f'$ with a question about $f$	derivative already drawn	average slope of $f$ from the graph	compare with a horizontal line at that height
"does the theorem guarantee"	a yes-or-no prompt	check hypotheses first	answer with the failed hypothesis

① Compute the average rate of change first and box it. Every later sentence refers to that number, and a wrong average makes the rest worthless.

② State continuity and differentiability with the correct interval types, and name the function the theorem is applied to. When the stem is about velocity or profit, the theorem is applied to that quantity, not to the abstract $f$ .

③ Write the conclusion as existence: "there is a $c$ in $(a,b)$ such that ...". If the stem then asks for the value, solve and filter.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = x^2$ on the interval $[0,4]$ . Find all values of $c$ guaranteed by the Mean Value Theorem.

Standard Execution:

Hypotheses. $f$ is a polynomial, so it is continuous on $[0,4]$ and differentiable on $(0,4)$ .

Average rate. $\frac{f(4) - f(0)}{4 - 0} = \frac{16 - 0}{4} = 4$ .

Equation. $f'(x) = 2x$ , so $2c = 4$ and $c = 2$ .

Filter. $2$ lies in $(0,4)$ , so it is the guaranteed value.

Exam Variation (Context / Parameter Shift): A differentiable function $f$ has $f(0) = 3$ and $f(4) = 11$ . Must there be a value $c$ in $(0,4)$ such that $f'(c) = 2$ ? Justify your answer.

Solution: Since $f$ is differentiable, it is continuous on $[0,4]$ and differentiable on $(0,4)$ . The average rate of change is $\frac{11 - 3}{4 - 0} = 2$ . By the Mean Value Theorem there is a $c$ in $(0,4)$ with $f'(c) = 2$ . The answer is yes, and the justification is the two hypotheses plus the computed average, not the number $2$ on its own.

Common Traps & Failure Signals:

Conclusion without hypotheses (Signal: "by the MVT there is a $c$ " as the first and only line): the two conditions were never stated, so the justification point is missing.
Average computed over the wrong interval (Signal: a denominator that does not match the stated endpoints): the endpoints of the table were read inconsistently.
Root kept outside the interval (Signal: a negative value of $c$ reported for an interval of positive numbers): the equation was solved but the answer was not filtered.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Average Skip: Compute the chord slope first.
Hypothesis Mumble: You wrote "continuous and differentiable" without attaching the correct interval to each word, so the sentence that was supposed to earn the justification point reads as a phrase you memorized rather than a condition you checked.
Existential Panic: The stem asked whether some $c$ must exist, you solved for a specific $c$ instead, and the answer now claims more than the theorem delivers.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released exams since 2017 contain a part that requires citing the Mean Value Theorem, and two of those five are the equal-endpoint case counted by the previous topic
10-Year Total Count: 5 of 9 Exams · 5 FRQ Parts; One part per exam, always a justification, and always scored on the hypotheses plus the conclusion
FRQ Years Appeared: 2018 (Q4b), 2019 (Q2a), 2021 (Q4d), 2023 (Q1b), 2025 (Q3B)

Practice Examples

MCQ 1

A Parabola on a Closed Interval

Let $f(x) = x^2$ on $[0,4]$ . Which value of $c$ does the Mean Value Theorem guarantee?

MCQ 2

When the Theorem Refuses

Let $f(x) = \frac{1}{x}$ on the interval $[-1, 1]$ . Which statement about the Mean Value Theorem is correct?

MCQ 3

Two Points and a Slope

A differentiable function $f$ satisfies $f(0) = 3$ and $f(4) = 11$ . Must there be a value $c$ in $(0,4)$ with $f'(c) = 2$ ?

FRQ 4

Justifying With Two Numbers

Let $f$ be a twice-differentiable function with $f(1) = 4$ and $f(5) = 16$ .

(a) Find the average rate of change of $f$ on the interval $[1,5]$ .

(b) Justify why there must exist a value $c$ in $(1,5)$ such that $f'(c) = 3$ .

(c) Let $g$ be the function defined by $g(x) = f(x) - 3x$ . Show that $g(1) = g(5)$ , and explain how Rolle's Theorem applied to $g$ proves the same conclusion as part (b).

(d) A student claims that there must also exist a value $d$ in $(1,5)$ with $f''(d) = 0$ . Determine whether this claim is justified by the given information.

Cheatsheet

Core Reference

	What to compute	What to conclude
"must there exist $c$ with $f'(c) = k$ "	average rate on $[a,b]$	yes if it equals $k$
"average velocity equals instantaneous"	$\frac{s(b)-s(a)}{b-a}$	some time inside has that velocity
"find the value guaranteed"	solve $f'(c) =$ average	keep roots inside $(a,b)$
"does the MVT guarantee"	check continuity and differentiability	yes or no, tied to the check
"must there be $c$ with $f''(c) = 0$ "	nothing new	not justified by the MVT

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 5.3 The Extreme Value Theorem (EVT)

Unit 5The Mean Value Theorem (MVT)

Concept 4.125Built for May 2027 AP Exam

The Mean Value Theorem (MVT)

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

Five of the nine released exams since 2017 contain a part that requires citing this theorem, and the wording is nearly identical each time.

What you’ll see: A table of values, a graph, or a differentiable function with two known points, followed by "must there exist a value $c$ ... such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ " or "explain why there must be at least one time at which the rate equals the average rate".
What you’ll do: Compute the average rate of change from the two endpoints, state the two hypotheses, and conclude that some interior point has that instantaneous rate.
Memory hook: Average somewhere becomes instant somewhere. The chord is copied by a tangent.
Use it on the exam: For MCQs: Solve $f'(c) = \frac{f(b)-f(a)}{b-a}$ directly and keep only the roots inside the open interval. Distractors come from roots outside it. For FRQs: The words "there must exist" are the instruction to justify rather than to solve. Write continuity, differentiability, and the average rate before the conclusion sentence.

Concept Review

The Secant Gets Copied Somewhere

Figure 1 — The secant across the interval has a slope; the theorem promises a tangent inside with the same slope. Rolle's theorem is the case where that slope is zero.

Mean Value Theorem: a tangent parallel to the secant exists inside the interval

The Statement and Its Two Halves

$f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \text{ in } (a,b)$

Rolle's Theorem Is the Level Case

The Sentence the Rubric Wants

What the stem asks	What the first line must be	What the last line must be
"must there exist $c$ with $f'(c) = k$ "	the average rate equals $k$	therefore such a $c$ exists
"explain why there must be a time"	$f$ is differentiable, hence continuous	by the MVT there is such a time
"does the MVT guarantee a value"	hypotheses hold or fail, with a reason	yes or no, tied to that reason
"find $c$ "	the average rate	solve the equation and keep roots inside

Common Misconception Check

Myth 1: "The theorem finds the location of $c$ ." Truth: It guarantees existence. Solving for $c$ is a separate request, and some functions make it impossible in closed form.
Myth 2: "Differentiability on the closed interval is what the theorem needs." Truth: Continuity is the closed-interval requirement and differentiability is the open-interval one. Swapping them is a written error the rubric notices.
Myth 3: "A function with a corner cannot satisfy the MVT." Truth: A corner can sit at an endpoint and the theorem still applies, because differentiability is only required inside the open interval.

Topic-Specific Study Method

Compute the Average, Then Claim the Instant

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table with two rows of data	$t$ and $f(t)$ columns	average rate from the two endpoints	compute, state hypotheses, conclude
"acceleration equals average acceleration"	velocity described in the stem	MVT applied to $v$	average rate of $v$ , then the claim
"average velocity" then "instantaneous"	two vocabulary words	the same theorem in motion language	one computation, one sentence
Graph of $f'$ with a question about $f$	derivative already drawn	average slope of $f$ from the graph	compare with a horizontal line at that height
"does the theorem guarantee"	a yes-or-no prompt	check hypotheses first	answer with the failed hypothesis

① Compute the average rate of change first and box it. Every later sentence refers to that number, and a wrong average makes the rest worthless.

③ Write the conclusion as existence: "there is a $c$ in $(a,b)$ such that ...". If the stem then asks for the value, solve and filter.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = x^2$ on the interval $[0,4]$ . Find all values of $c$ guaranteed by the Mean Value Theorem.

Standard Execution:

Hypotheses. $f$ is a polynomial, so it is continuous on $[0,4]$ and differentiable on $(0,4)$ .

Average rate. $\frac{f(4) - f(0)}{4 - 0} = \frac{16 - 0}{4} = 4$ .

Equation. $f'(x) = 2x$ , so $2c = 4$ and $c = 2$ .

Filter. $2$ lies in $(0,4)$ , so it is the guaranteed value.

Exam Variation (Context / Parameter Shift): A differentiable function $f$ has $f(0) = 3$ and $f(4) = 11$ . Must there be a value $c$ in $(0,4)$ such that $f'(c) = 2$ ? Justify your answer.

Solution: Since $f$ is differentiable, it is continuous on $[0,4]$ and differentiable on $(0,4)$ . The average rate of change is $\frac{11 - 3}{4 - 0} = 2$ . By the Mean Value Theorem there is a $c$ in $(0,4)$ with $f'(c) = 2$ . The answer is yes, and the justification is the two hypotheses plus the computed average, not the number $2$ on its own.

Common Traps & Failure Signals:

Conclusion without hypotheses (Signal: "by the MVT there is a $c$ " as the first and only line): the two conditions were never stated, so the justification point is missing.
Average computed over the wrong interval (Signal: a denominator that does not match the stated endpoints): the endpoints of the table were read inconsistently.
Root kept outside the interval (Signal: a negative value of $c$ reported for an interval of positive numbers): the equation was solved but the answer was not filtered.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Average Skip: Compute the chord slope first.
Hypothesis Mumble: You wrote "continuous and differentiable" without attaching the correct interval to each word, so the sentence that was supposed to earn the justification point reads as a phrase you memorized rather than a condition you checked.
Existential Panic: The stem asked whether some $c$ must exist, you solved for a specific $c$ instead, and the answer now claims more than the theorem delivers.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released exams since 2017 contain a part that requires citing the Mean Value Theorem, and two of those five are the equal-endpoint case counted by the previous topic
10-Year Total Count: 5 of 9 Exams · 5 FRQ Parts; One part per exam, always a justification, and always scored on the hypotheses plus the conclusion
FRQ Years Appeared: 2018 (Q4b), 2019 (Q2a), 2021 (Q4d), 2023 (Q1b), 2025 (Q3B)

Practice Examples

MCQ 1

A Parabola on a Closed Interval

Let $f(x) = x^2$ on $[0,4]$ . Which value of $c$ does the Mean Value Theorem guarantee?

MCQ 2

When the Theorem Refuses

Let $f(x) = \frac{1}{x}$ on the interval $[-1, 1]$ . Which statement about the Mean Value Theorem is correct?

MCQ 3

Two Points and a Slope

A differentiable function $f$ satisfies $f(0) = 3$ and $f(4) = 11$ . Must there be a value $c$ in $(0,4)$ with $f'(c) = 2$ ?

FRQ 4

Justifying With Two Numbers

Let $f$ be a twice-differentiable function with $f(1) = 4$ and $f(5) = 16$ .

(a) Find the average rate of change of $f$ on the interval $[1,5]$ .

(b) Justify why there must exist a value $c$ in $(1,5)$ such that $f'(c) = 3$ .

(c) Let $g$ be the function defined by $g(x) = f(x) - 3x$ . Show that $g(1) = g(5)$ , and explain how Rolle's Theorem applied to $g$ proves the same conclusion as part (b).

(d) A student claims that there must also exist a value $d$ in $(1,5)$ with $f''(d) = 0$ . Determine whether this claim is justified by the given information.

Cheatsheet

Core Reference

	What to compute	What to conclude
"must there exist $c$ with $f'(c) = k$ "	average rate on $[a,b]$	yes if it equals $k$
"average velocity equals instantaneous"	$\frac{s(b)-s(a)}{b-a}$	some time inside has that velocity
"find the value guaranteed"	solve $f'(c) =$ average	keep roots inside $(a,b)$
"does the MVT guarantee"	check continuity and differentiability	yes or no, tied to the check
"must there be $c$ with $f''(c) = 0$ "	nothing new	not justified by the MVT

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 5.3 The Extreme Value Theorem (EVT)

Unit 5The Mean Value Theorem (MVT)

Concept 4.125Built for May 2027 AP Exam

The Mean Value Theorem (MVT)

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

Five of the nine released exams since 2017 contain a part that requires citing this theorem, and the wording is nearly identical each time.

What you’ll see: A table of values, a graph, or a differentiable function with two known points, followed by "must there exist a value $c$ ... such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ " or "explain why there must be at least one time at which the rate equals the average rate".
What you’ll do: Compute the average rate of change from the two endpoints, state the two hypotheses, and conclude that some interior point has that instantaneous rate.
Memory hook: Average somewhere becomes instant somewhere. The chord is copied by a tangent.
Use it on the exam: For MCQs: Solve $f'(c) = \frac{f(b)-f(a)}{b-a}$ directly and keep only the roots inside the open interval. Distractors come from roots outside it. For FRQs: The words "there must exist" are the instruction to justify rather than to solve. Write continuity, differentiability, and the average rate before the conclusion sentence.

Concept Review

The Secant Gets Copied Somewhere

Figure 1 — The secant across the interval has a slope; the theorem promises a tangent inside with the same slope. Rolle's theorem is the case where that slope is zero.

Mean Value Theorem: a tangent parallel to the secant exists inside the interval

The Statement and Its Two Halves

$f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \text{ in } (a,b)$

Rolle's Theorem Is the Level Case

The Sentence the Rubric Wants

What the stem asks	What the first line must be	What the last line must be
"must there exist $c$ with $f'(c) = k$ "	the average rate equals $k$	therefore such a $c$ exists
"explain why there must be a time"	$f$ is differentiable, hence continuous	by the MVT there is such a time
"does the MVT guarantee a value"	hypotheses hold or fail, with a reason	yes or no, tied to that reason
"find $c$ "	the average rate	solve the equation and keep roots inside

Common Misconception Check

Myth 1: "The theorem finds the location of $c$ ." Truth: It guarantees existence. Solving for $c$ is a separate request, and some functions make it impossible in closed form.
Myth 2: "Differentiability on the closed interval is what the theorem needs." Truth: Continuity is the closed-interval requirement and differentiability is the open-interval one. Swapping them is a written error the rubric notices.
Myth 3: "A function with a corner cannot satisfy the MVT." Truth: A corner can sit at an endpoint and the theorem still applies, because differentiability is only required inside the open interval.

Topic-Specific Study Method

Compute the Average, Then Claim the Instant

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table with two rows of data	$t$ and $f(t)$ columns	average rate from the two endpoints	compute, state hypotheses, conclude
"acceleration equals average acceleration"	velocity described in the stem	MVT applied to $v$	average rate of $v$ , then the claim
"average velocity" then "instantaneous"	two vocabulary words	the same theorem in motion language	one computation, one sentence
Graph of $f'$ with a question about $f$	derivative already drawn	average slope of $f$ from the graph	compare with a horizontal line at that height
"does the theorem guarantee"	a yes-or-no prompt	check hypotheses first	answer with the failed hypothesis

① Compute the average rate of change first and box it. Every later sentence refers to that number, and a wrong average makes the rest worthless.

③ Write the conclusion as existence: "there is a $c$ in $(a,b)$ such that ...". If the stem then asks for the value, solve and filter.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = x^2$ on the interval $[0,4]$ . Find all values of $c$ guaranteed by the Mean Value Theorem.

Standard Execution:

Hypotheses. $f$ is a polynomial, so it is continuous on $[0,4]$ and differentiable on $(0,4)$ .

Average rate. $\frac{f(4) - f(0)}{4 - 0} = \frac{16 - 0}{4} = 4$ .

Equation. $f'(x) = 2x$ , so $2c = 4$ and $c = 2$ .

Filter. $2$ lies in $(0,4)$ , so it is the guaranteed value.

Exam Variation (Context / Parameter Shift): A differentiable function $f$ has $f(0) = 3$ and $f(4) = 11$ . Must there be a value $c$ in $(0,4)$ such that $f'(c) = 2$ ? Justify your answer.

Solution: Since $f$ is differentiable, it is continuous on $[0,4]$ and differentiable on $(0,4)$ . The average rate of change is $\frac{11 - 3}{4 - 0} = 2$ . By the Mean Value Theorem there is a $c$ in $(0,4)$ with $f'(c) = 2$ . The answer is yes, and the justification is the two hypotheses plus the computed average, not the number $2$ on its own.

Common Traps & Failure Signals:

Conclusion without hypotheses (Signal: "by the MVT there is a $c$ " as the first and only line): the two conditions were never stated, so the justification point is missing.
Average computed over the wrong interval (Signal: a denominator that does not match the stated endpoints): the endpoints of the table were read inconsistently.
Root kept outside the interval (Signal: a negative value of $c$ reported for an interval of positive numbers): the equation was solved but the answer was not filtered.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Average Skip: Compute the chord slope first.
Hypothesis Mumble: You wrote "continuous and differentiable" without attaching the correct interval to each word, so the sentence that was supposed to earn the justification point reads as a phrase you memorized rather than a condition you checked.
Existential Panic: The stem asked whether some $c$ must exist, you solved for a specific $c$ instead, and the answer now claims more than the theorem delivers.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released exams since 2017 contain a part that requires citing the Mean Value Theorem, and two of those five are the equal-endpoint case counted by the previous topic
10-Year Total Count: 5 of 9 Exams · 5 FRQ Parts; One part per exam, always a justification, and always scored on the hypotheses plus the conclusion
FRQ Years Appeared: 2018 (Q4b), 2019 (Q2a), 2021 (Q4d), 2023 (Q1b), 2025 (Q3B)

Practice Examples

MCQ 1

A Parabola on a Closed Interval

Let $f(x) = x^2$ on $[0,4]$ . Which value of $c$ does the Mean Value Theorem guarantee?

MCQ 2

When the Theorem Refuses

Let $f(x) = \frac{1}{x}$ on the interval $[-1, 1]$ . Which statement about the Mean Value Theorem is correct?

MCQ 3

Two Points and a Slope

A differentiable function $f$ satisfies $f(0) = 3$ and $f(4) = 11$ . Must there be a value $c$ in $(0,4)$ with $f'(c) = 2$ ?

FRQ 4

Justifying With Two Numbers

Let $f$ be a twice-differentiable function with $f(1) = 4$ and $f(5) = 16$ .

(a) Find the average rate of change of $f$ on the interval $[1,5]$ .

(b) Justify why there must exist a value $c$ in $(1,5)$ such that $f'(c) = 3$ .

(c) Let $g$ be the function defined by $g(x) = f(x) - 3x$ . Show that $g(1) = g(5)$ , and explain how Rolle's Theorem applied to $g$ proves the same conclusion as part (b).

(d) A student claims that there must also exist a value $d$ in $(1,5)$ with $f''(d) = 0$ . Determine whether this claim is justified by the given information.

Cheatsheet

Core Reference

	What to compute	What to conclude
"must there exist $c$ with $f'(c) = k$ "	average rate on $[a,b]$	yes if it equals $k$
"average velocity equals instantaneous"	$\frac{s(b)-s(a)}{b-a}$	some time inside has that velocity
"find the value guaranteed"	solve $f'(c) =$ average	keep roots inside $(a,b)$
"does the MVT guarantee"	check continuity and differentiability	yes or no, tied to the check
"must there be $c$ with $f''(c) = 0$ "	nothing new	not justified by the MVT

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 5.3 The Extreme Value Theorem (EVT)