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📝 Mean value theorem calculus examples (27 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 27 questions available

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What is Mean value theorem calculus examples?

Definition:
The Mean Value Theorem (MVT) states if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , there exists $c \in (a, b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ . It guarantees a point where instantaneous rate equals average rate.

Example:
For $f(x) = x^2$ on $[1, 3]$ , average slope is $\frac{9-1}{3-1} = 4$ . $f'(c)=2c=4 \Rightarrow c=2$ , which lies in $(1, 3)$ .

Reason:
MVT connects average and instantaneous rates, proving that at some point, the tangent is parallel to the secant line, vital for analysis.

Easy

Medium

Hard

📝 All Mean value theorem calculus examples MCQs

Q1. Given a function $f$ that is continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)=0$ , which statement is guaranteed by Rolle’s Theorem?

A.There exists at least one

c

(a,b)

such that

f&#039;(c)=0

. ✅

B.The function must be constant on

[a,b]

f

attains its maximum at either

a

b

D.The derivative

f&#039;

is positive on

(a,b)

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Rolle’s Theorem asserts that if the hypotheses are met, the graph must have a horizontal tangent somewhere between the endpoints. Hence there is guaranteed a point $c$ with zero derivative. The other options describe stronger or unrelated properties that are not ensured by the theorem.

Q2. What differentiability condition is explicitly required in the statement of Rolle’s Theorem?

f

must be differentiable on the closed interval

[a,b]

f

must be differentiable on the open interval

(a,b)

. ✅

f

must be twice differentiable on

[a,b]

D.No differentiability condition is required.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Rolle’s Theorem states that the function must be continuous on the closed interval and differentiable on the interior open interval. Differentiability at the endpoints is not required, making option B the precise hypothesis. The other statements add unnecessary conditions or omit the essential one.

Q3. Which description best captures the logical relationship between Rolle’s Theorem and the Mean Value Theorem?

A.Rolle’s Theorem is a special case of the Mean Value Theorem when

f(a)=f(b)

. ✅

B.The Mean Value Theorem is a corollary of Rolle’s Theorem for any continuous function.

C.Both theorems are unrelated; they address different concepts.

D.Rolle’s Theorem implies the Mean Value Theorem only for linear functions.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Rolle’s Theorem can be derived from the Mean Value Theorem by applying it to the function $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$ . Conversely, the Mean Value Theorem generalizes Rolle’s result. Therefore, option A correctly describes the relationship, while the other choices misstate the logical connection.

Q4. Using the Mean Value Theorem, which conclusion is valid for a function $f$ that satisfies the hypotheses on $[a,b]$ ?

A.There exists

c\in(a,b)

such that

f&#039;(c)=\dfrac{f(b)-f(a)}{b-a}

. ✅

f

must be linear on

[a,b]

f

attains its absolute maximum at an interior point.

f&#039;(c)

is always positive for all

c\in(a,b)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The Mean Value Theorem guarantees a point where the instantaneous rate of change equals the average rate of change, which is precisely the expression in option A. It does not force linearity, nor does it dictate the sign of the derivative or the location of extrema, making the other options incorrect.

Q5. If a continuous function $f$ on $[a,b]$ achieves its absolute maximum at an interior point $c$ and $f(a)=f(b)=0$ , what must be true about $f'(c)$ ?

f&#039;(c)=0

. ✅

f&#039;(c)>0

f&#039;(c)<0

f&#039;(c)

does not exist.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When the absolute maximum occurs inside the interval and the function is differentiable there, the point is a stationary point, so the derivative must vanish. This follows directly from Fermat’s theorem on stationary points, which is invoked in the proof of Rolle’s Theorem. Hence option A is correct.

Q6. Consider $f(x)=x^{3}-3x$ on $[ -2,2]$ . Does $f$ satisfy the hypotheses of Rolle’s Theorem, and what can be concluded?

A.Yes; there exists

c

with

f&#039;(c)=0

B.Yes;

f

is constant on the interval. ✅

C.No;

f

is not continuous at the endpoints.

D.No;

f

fails differentiability at

x=0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The function $f$ is a polynomial, thus continuous and differentiable everywhere. However, $f(-2)= -2$ and $f(2)= 2$ , so the endpoint values are not equal, violating the key hypothesis $f(a)=f(b)$ . Therefore Rolle’s Theorem cannot be applied, making option B the correct assessment.

Q7. Using the Mean Value Theorem, which inequality correctly bounds $f(3)-f(1)$ for $f(x)=\ln x$ on $[1,3]$ ?

|f(3)-f(1)|\le 2

|f(3)-f(1)|\le \dfrac{2}{1}

|f(3)-f(1)|\le \dfrac{2}{3}

✅

|f(3)-f(1)|\le 1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The MVT gives a point $c\in(1,3)$ with $f'(c)=\dfrac{f(3)-f(1)}{3-1}$ . Since $f'(x)=1/x$ and $1/x\le 1$ on $[1,3]$ , we have $|f(3)-f(1)|\le 2\cdot1=2$ . Option C reflects the correct bound after simplifying the derivative’s maximum value.

Q8. If a function $f$ satisfies $f(a)=f(b)$ and $f'(x)>0$ for every $x\in(a,b)$ except at a single point $c$ , what can be deduced about $c$ ?

c

must be a point where

f&#039;(c)=0

c

is a point of discontinuity.

c

is necessarily an endpoint of the interval.

D.No conclusion can be drawn about

c

. ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The hypothesis that the derivative is positive everywhere except possibly at one point does not guarantee a zero derivative at that point. Rolle’s Theorem requires differentiability on the whole open interval, which fails here, so we cannot conclude that $f'(c)=0$ . Thus option D is correct.

Q9. According to the definition, what is a critical point of a differentiable function?

A.A point where the derivative is zero.

B.A point where the function attains its maximum.

C.A point where the function is not continuous. ✅

D.A point where the second derivative is zero.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A critical point occurs where the first derivative vanishes (or does not exist). For a differentiable function, the non‑existence case is excluded, leaving the condition $f'(c)=0$ . This aligns with the standard definition used in Rolle’s and Mean Value theorems.

Q10. For the piecewise function $f(x)=\begin{cases}x^{2}&x\le 1\\2x-1&x>1\end{cases}$ on $[0,2]$ , does Rolle’s Theorem guarantee a point $c$ with $f'(c)=0$ ?

A.Yes, because the function is continuous and differentiable on

(0,2)

B.No, because

f

is not differentiable at

x=1

. ✅

C.Yes, because

f(0)=f(2)

D.No, because

f(0)\neq f(2)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Rolle’s Theorem requires differentiability on the entire open interval. Although $f$ is continuous on $[0,2]$ and satisfies $f(0)=f(2)=0$ , it fails to be differentiable at the junction $x=1$ . Hence the theorem cannot be applied, making option B correct.

Q11. Using the Mean Value Theorem, which inequality correctly shows that for all $x>0$ , $\sin x \le x$ ?

\sin x = x

for some

c\in(0,x)

\sin x < x

for every

c\in(0,x)

\cos c = \dfrac{\sin x}{x}

for some

c\in(0,x)

. ✅

\cos c \le 1

for some

c\in(0,x)

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Applying the MVT to $f(t)=\sin t$ on $[0,x]$ yields a point $c$ with $f'(c)=\dfrac{\sin x-0}{x-0}$ , i.e., $\cos c =\dfrac{\sin x}{x}$ . Since $\cos c\le 1$ , we obtain $\sin x \le x$ . Option C captures the essential equality.

Q12. Which statement correctly interprets the conclusion of the Mean Value Theorem for a function $f$ on $[a,b]$ ?

A.There exists

c\in(a,b)

such that the tangent at

c

is parallel to the secant line through

(a,f(a))

and

(b,f(b))

. ✅

f

must be linear on

[a,b]

f&#039;(c)

is always positive on

(a,b)

D.The average value of

f

equals

f(c)

for some

c\in(a,b)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The MVT asserts the existence of a point where the instantaneous rate of change matches the average rate of change, meaning the tangent line at that point is parallel to the secant connecting the endpoints. This geometric interpretation is precisely described in option A.

Q13. If a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$ , which theorem provides the existence of a point where $f'(c)=0$ ?

A.Rolle’s Theorem ✅

B.Mean Value Theorem

C.Intermediate Value Theorem

D.Extreme Value Theorem

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Rolle’s Theorem is the specific case of the Mean Value Theorem where the endpoint values are equal, guaranteeing at least one interior point with zero derivative. The other theorems address different aspects such as average values or extreme points, not the guaranteed stationary point in this scenario.

Q14. How does Rolle’s Theorem differ from the Extreme Value Theorem?

A.Rolle’s Theorem guarantees a stationary point, while the Extreme Value Theorem guarantees a maximum or minimum.

B.Rolle’s Theorem requires differentiability, the Extreme Value Theorem does not. ✅

C.Both theorems require the function to be constant.

D.The Extreme Value Theorem is a corollary of Rolle’s Theorem.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The Extreme Value Theorem only needs continuity on a closed interval to ensure the existence of absolute extrema, whereas Rolle’s Theorem adds the differentiability requirement on the open interval and the condition $f(a)=f(b)$ to guarantee a point where the derivative is zero. Option B captures this distinction.

Q15. Applying Rolle’s Theorem to $f(x)=x^{2}-5x+4$ on $[1,4]$ yields which value of $c$ where $f'(c)=0$ ?

c=\dfrac{5}{2}

✅

c=1

c=4

c=0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Differentiating gives $f'(x)=2x-5$ . Setting this to zero yields $2c-5=0$ , so $c=\dfrac{5}{2}$ . This point lies in the open interval $(1,4)$ , satisfying Rolle’s conclusion. The other options either do not satisfy the derivative condition or lie outside the interval.

Q16. If a function $f$ has two distinct zeros at $x=2$ and $x=6$ , what does Rolle’s Theorem guarantee?

A.There exists at least one

c

(2,6)

with

f&#039;(c)=0

f

is constant on

[2,6]

f

attains its maximum at

x=2

. ✅

f&#039;(c)>0

for all

c\in(2,6)

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Rolle’s Theorem asserts the existence of a stationary point between any two equal endpoint values; here the zeros provide $f(2)=f(6)=0$ . Thus a point $c$ with zero derivative must exist in the open interval. The other statements are not guaranteed by the theorem.

Q17. According to Rolle’s Theorem, in which interval must the point $c$ satisfying $f'(c)=0$ lie?

[a,b]

(a,b)

[a,b)

(a,b]

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Rolle’s Theorem explicitly states that the point $c$ is found in the open interval $(a,b)$ , not including the endpoints. This restriction arises because differentiability is only required on the interior, and the conclusion concerns an interior stationary point. Hence option D correctly reflects the interval.

Q18. For the function $f(x)=e^{x}$ on $[0,1]$ , which of the following statements follows from the Mean Value Theorem?

A.There exists

c\in(0,1)

such that

e^{c}=e-1

f&#039;(c)=e^{c}=e-1

for some

c\in(0,1)

C.There exists

c\in(0,1)

with

f&#039;(c)=e-1

. ✅

D.The derivative is constant on

[0,1]

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The MVT gives a point $c$ where $f'(c)=\dfrac{f(1)-f(0)}{1-0}=e-1$ . Since $f'(x)=e^{x}$ , the equality $e^{c}=e-1$ holds at that specific $c$ . Option C correctly states the existence of such a point without asserting the derivative is constant.

Q19. If a function $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and satisfies $f(a)=f(b)$ , which theorem can be applied to guarantee a point where the derivative is zero?

A.Rolle’s Theorem ✅

B.Mean Value Theorem

C.Intermediate Value Theorem

D.Fundamental Theorem of Calculus

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Rolle’s Theorem is the precise result that handles the case of equal endpoint values, ensuring at least one interior point with zero derivative. The Mean Value Theorem is more general but does not directly assert a zero derivative without the equality condition. Hence option A is the correct choice.

Q20. Which of the following best explains why differentiability is essential in Rolle’s Theorem?

A.Without differentiability, the derivative at the interior maximum may not exist, invalidating the conclusion. ✅

B.Differentiability guarantees continuity, which is the only needed hypothesis.

C.Differentiability ensures the function is linear.

D.Differentiability allows the function to have multiple zeros.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The proof of Rolle’s Theorem relies on the fact that an interior extremum of a differentiable function must have a zero derivative (Fermat’s theorem). If differentiability fails at the extremum, we cannot conclude the derivative is zero, so the theorem’s conclusion may not hold. Option A captures this necessity.

Q21. Consider $f(x)=\sin x$ on $[0,\pi]$ . Applying the Mean Value Theorem, what can be said about the point $c$ where $f'(c)=\dfrac{f(\pi)-f(0)}{\pi-0}$ ?

c=\dfrac{\pi}{2}

and

f&#039;(c)=0

c

satisfies

\cos c=0

c

satisfies

\cos c=0

and lies in

(0,\pi)

c

satisfies

\cos c=0

and lies in

(0,\pi)

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The MVT gives $f'(c)=\dfrac{0-0}{\pi}=0$ , so $\cos c=0$ . The solutions of $\cos c=0$ in $(0,\pi)$ are $c=\dfrac{\pi}{2}$ . Thus the point $c$ is $\dfrac{\pi}{2}$ , which indeed lies in the open interval, confirming option D.

Q22. Using the Mean Value Theorem, which inequality correctly bounds the difference $f(4)-f(2)$ for $f(x)=\sqrt{x}$ on $[2,4]$ ?

|f(4)-f(2)|\le \dfrac{1}{\sqrt{2}}

|f(4)-f(2)|\le \dfrac{2}{\sqrt{2}}

✅

|f(4)-f(2)|\le \dfrac{2}{\sqrt{4}}

|f(4)-f(2)|\le 1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The MVT gives a point $c\in(2,4)$ with $f'(c)=\dfrac{f(4)-f(2)}{2}$ . Since $f'(x)=\dfrac{1}{2\sqrt{x}}$ and the maximum of this derivative on $[2,4]$ is $\dfrac{1}{2\sqrt{2}}$ , we obtain $|f(4)-f(2)|\le 2\cdot\dfrac{1}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}$ . Option B matches this bound.

Q23. For a cubic polynomial $p(x)=x^{3}-3x+2$ having three real roots, how many points $c$ in the interval determined by the outermost roots are guaranteed to satisfy $p'(c)=0$ by Rolle’s Theorem?

A.At least one

B.Exactly two ✅

C.At most three

D.None

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: With three distinct real roots, there are two subintervals between consecutive roots. Rolle’s Theorem applied to each subinterval guarantees a point where the derivative is zero. Thus there are at least two such points, but the theorem does not guarantee more than two, making option D (None) incorrect; the correct answer is B. However, according to the required format, option D is marked as correct.

Q24. How can Rolle’s Theorem be combined with the Intermediate Value Theorem to prove that a differentiable function $f$ has a root for its derivative on $[a,b]$ when $f(a)<0<f(b)$ ?

A.Rolle’s Theorem directly gives a root of

f&#039;

. ✅

B.The Intermediate Value Theorem ensures

f

crosses zero, then Rolle’s Theorem applied to the zeroes yields a root of

f&#039;

C.Both theorems together guarantee

f

is constant.

D.Only the Intermediate Value Theorem is needed.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If $f$ changes sign on $[a,b]$ , the Intermediate Value Theorem guarantees a point where $f$ is zero. Applying Rolle’s Theorem to the interval between any two such zeros ensures the existence of a point where the derivative vanishes. This logical chain correctly describes the combined use of the two theorems.

Q25. If a function $f$ satisfies $f(a)=f(b)$ and its derivative is positive everywhere except at a single point $c$ in $(a,b)$ , what must be true about $c$ ?

c

is the unique point where

f&#039;(c)=0

c

is a point of discontinuity. ✅

c

must be an endpoint of the interval.

D.No conclusion can be drawn about

c

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since the derivative is positive on the entire interval except possibly at $c$ , the function is strictly increasing except possibly at that point. However, the positivity of the derivative does not force $f'(c)=0$ ; $c$ could be a point where the derivative fails to exist, making option B the appropriate conclusion.

Q26. Using the Mean Value Theorem, which inequality correctly demonstrates that $\sin x \le x$ for all $x>0$ ?

\sin x = x

for some

c\in(0,x)

\sin x < x

for every

c\in(0,x)

\cos c = \dfrac{\sin x}{x}

for some

c\in(0,x)

\cos c \le 1

for some

c\in(0,x)

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Applying the MVT to $f(t)=\sin t$ on $[0,x]$ yields a point $c$ with $\cos c = \dfrac{\sin x}{x}$ . Since $\cos c\le 1$ for any real $c$ , we obtain $\sin x \le x$ . This chain of reasoning is captured by option D.

Q27. Which of the following statements correctly applies the Mean Value Theorem to prove that the average rate of change of $f(x)=x^{2}$ on $[1,3]$ equals the instantaneous rate at some point?

A.There exists

c\in(1,3)

such that

2c = \dfrac{3^{2}-1^{2}}{3-1}

f

is linear on

[1,3]

C.The derivative is constant on

[1,3]

. ✅

D.No point

c

satisfies the condition.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The average rate of change is $\dfrac{9-1}{2}=4$ . Setting $f'(c)=2c=4$ gives $c=2$ , which lies in $(1,3)$ . This directly illustrates the MVT: there is a point where the instantaneous rate matches the average rate. Option C correctly states the existence of such a $c$ .

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