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📝 Mean value theorem applications (23 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 23 questions available

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What is Mean value theorem applications?

Definition:
MVT applications include proving inequalities, bounding errors, and establishing function properties. For instance, if $|f'(x)| \le M$ , then $|f(b)-f(a)| \le M|b-a|$ . It also proves that functions with zero derivative are constant, linking derivatives to function behavior.

Example:
To show $|\sin(b) - \sin(a)| \le |b-a|$ , use MVT on $\sin(x)$ where $|f'(c)| = |\cos(c)| \le 1$ , so difference $\le 1 \cdot |b-a|$ .

Reason:
It transforms local derivative information into global function bounds, enabling rigorous proofs of stability and continuity in mathematical analysis.

Easy

Medium

Hard

📝 All Mean value theorem applications MCQs

Q1. Using the Mean‑Value Theorem, if $f'(x)>0$ for every $x\in(a,b)$ and $x_{1}<x_{2}$ , what can we infer about the values of $f(x_{1})$ and $f(x_{2})$ ?

f(x_{1})>f(x_{2})

f(x_{1})=f(x_{2})

f(x_{1})<f(x_{2})

✅

D.Cannot determine from the information given

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: By the Mean‑Value Theorem there exists $c\in(x_{1},x_{2})$ with $f'(c)=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$ . Since $f'(c)>0$ and $x_{2}-x_{1}>0$ , the numerator must be positive, giving $f(x_{2})>f(x_{1})$ . Hence option C is correct.

Q2. Compare the statements: (i) If $f'(x)>0$ on $(a,b)$ then $f$ is increasing on $[a,b]$ . (ii) If $f$ is increasing on $[a,b]$ then $f'(x)>0$ on $(a,b)$ . Which combination is true?

A.Both statements are true

B.Statement (i) true, statement (ii) false ✅

C.Statement (i) false, statement (ii) true

D.Both statements are false

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The Mean‑Value Theorem guarantees that a positive derivative forces the function to rise, making (i) true. The converse need not hold because a function can be increasing while its derivative is zero at some points, so (ii) is false. Therefore option B is correct.

Q3. If a function $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and satisfies $f'(x)=0$ for every $x\in(a,b)$ , what does Theorem 4.1.2(c) assert?

f

is constant on

[a,b]

✅

f

is increasing on

[a,b]

f

is decreasing on

[a,b]

f

attains a maximum at some interior point

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Part (c) of Theorem 4.1.2 follows directly from the Mean‑Value Theorem: applying the theorem to any subinterval yields a point where the derivative equals the average change, which is zero. Hence the function cannot change value, implying it is constant on the whole interval. Option A captures this conclusion.

Q4. Suppose $f'(c)=5$ for some $c\in( x_{1},x_{2})$ and $x_{2}-x_{1}=2$ . What is $f(x_{2})-f(x_{1})$ according to the Mean‑Value Theorem?

A.10 ✅

B.5

C.2

D.0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The Mean‑Value Theorem gives $f(x_{2})-f(x_{1})=f'(c)(x_{2}-x_{1})$ . Substituting $f'(c)=5$ and $x_{2}-x_{1}=2$ yields $5\times2=10$ . Thus the correct value is 10, corresponding to option A.

Q5. Let $g(x)=f(x)-x$ where $f'(x)>0$ on $(a,b)$ . Which statement about the monotonicity of $g$ is guaranteed?

g

is increasing

g

is decreasing

g

is constant

D.Cannot be determined from the given information ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The derivative of $g$ is $g'(x)=f'(x)-1$ . While we know $f'(x)>0$ , we do not know whether it is larger or smaller than 1, so the sign of $g'(x)$ is indeterminate. Hence no monotonicity can be guaranteed, making option D correct.

Q6. How does the Mean‑Value Theorem lead to the conclusion that a function with zero derivative everywhere on $(a,b)$ must be constant on $[a,b]$ ?

A.By contradiction

B.By applying the theorem to any subinterval ✅

C.By integrating the derivative

D.By using the Intermediate Value Theorem

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Applying the Mean‑Value Theorem to an arbitrary subinterval $[x_{1},x_{2}]$ yields a point $c$ with $f'(c)=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$ . Since $f'(c)=0$ , the numerator must be zero, forcing $f(x_{2})=f(x_{1})$ . Repeating this argument shows the function is constant throughout the interval. Hence option B is correct.

Q7. If $f$ is continuous on $[0,2]$ , differentiable on $(0,2)$ , and $f'(x)=3$ for all $x$ , what is $f(2)-f(0)$ ?

A.6 ✅

B.3

C.9

D.Cannot be determined without knowing

f(0)

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: With a constant derivative, the Mean‑Value Theorem gives $f(2)-f(0)=f'(c)(2-0)=3\times2=6$ . The actual values of $f$ at the endpoints are irrelevant; the difference is determined solely by the derivative and interval length. Thus option A is correct.

Q8. Given a continuous function on $[a,b]$ that is differentiable on $(a,b)$ and satisfies $f(b)-f(a)=0$ , what does the Mean‑Value Theorem guarantee?

A.There exists

c

with \(f'(c)=0 ✅

B.There exists

c

with

f&#039;(c)>0

C.There exists

c

with

f&#039;(c)<0

D.No guarantee about any derivative value

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The Mean‑Value Theorem asserts the existence of a point $c\in(a,b)$ where $f'(c)=\frac{f(b)-f(a)}{b-a}$ . Since the numerator is zero, the quotient is zero, so $f'(c)=0$ . This is exactly the statement of Rolle’s Theorem, making option A correct.

Q9. Consider $f(x)=x^{3}$ and $g(x)=x^{3}+x$ on $[0,1]$ . Both satisfy the hypotheses of the Mean‑Value Theorem. Which comparison of their derivatives at the guaranteed points $c_{f}$ and $c_{g}$ is correct?

A.Both have equal derivatives at their respective points

f&#039;

c_{f}

is larger than

g&#039;

c_{g}

g&#039;

c_{g}

is larger than

f&#039;

c_{f}

✅

D.Derivatives are unrelated

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $f$ , the average slope is $(1-0)/(1-0)=1$ , giving $c_{f}=\sqrt{1/3}$ and $f'(c_{f})=1$ . For $g$ , the average slope is $(1+1-0)/(1-0)=2$ , leading to the same $c_{g}=\sqrt{1/3}$ but $g'(c_{g})=3c_{g}^{2}+1=2$ . Hence the derivative of $g$ at its point is larger, so option C is correct.

Q10. Why is the strict condition $f'(x)>0$ essential for concluding that $f$ is increasing, rather than the weaker condition $f'(x)\ge0$ ?

A.Because equality could allow constant segments ✅

B.Because the derivative might be zero at isolated points

C.Because the Mean‑Value Theorem requires strict positivity

D.Because continuity fails otherwise

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If $f'(x)\ge0$ everywhere, the function could be flat on intervals, producing no increase despite the derivative never being negative. The strict inequality guarantees a positive change over any subinterval, ensuring genuine increase. Thus option A correctly captures why strict positivity is required.

Q11. If a function $h$ has a point $c\in(a,b)$ with $h'(c)=0$ , which of the following must be true?

h

has a local maximum at

c

h

has a local minimum at

c

h

is constant on

[a,b]

D.None of the above necessarily holds ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: A zero derivative at a point is a necessary condition for a local extremum but not sufficient; the function could be increasing, decreasing, or have an inflection point there. Therefore none of the listed conclusions is guaranteed, making option D correct.

Q12. Using the Mean‑Value Theorem, show that if two functions have the same derivative on $(a,b)$ , then they differ by a constant on $[a,b]$ .

A.By integrating the difference of the functions

B.By applying the theorem to their difference ✅

C.By Rolle’s Theorem

D.By the Intermediate Value Theorem

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Define $k(x)=f(x)-g(x)$ . Since $k'(x)=f'(x)-g'(x)=0$ on $(a,b)$ , applying the Mean‑Value (or Rolle’s) Theorem to $k$ on any subinterval yields $k$ constant there. Extending over the whole interval shows $f$ and $g$ differ by a constant. Thus option B is correct.

Q13. If $f$ is continuous on $[1,4]$ , differentiable on $(1,4)$ , and $f(4)-f(1)=9$ , what does the Mean‑Value Theorem guarantee?

A.There exists

c

with

f&#039;(c)=3

✅

B.There exists

c

with

f&#039;(c)=9

C.There exists

c

with

f&#039;(c)=0

D.No guarantee about any derivative value

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average slope over $[1,4]$ is $\frac{9}{4-1}=3$ . The Mean‑Value Theorem ensures a point $c$ where the instantaneous slope equals this average, i.e., $f'(c)=3$ . Hence option A is correct.

Q14. Compare the conclusions of Theorem 4.1.2 parts (a) and (c). Which statement is stronger?

A.Part (a) is stronger because it gives monotonicity ✅

B.Part (c) is stronger because it gives constancy

C.Both are incomparable in strength

D.They are equivalent statements

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Part (a) asserts that a positive derivative forces the function to be increasing, a property that implies but does not require constancy. Part (c) only concludes constancy when the derivative is identically zero. Thus the monotonicity conclusion of part (a) is the stronger result. Option A is correct.

Q15. If $f'$ changes sign from positive to negative at a point $d$ , what can we infer about $f$ at $d$ using the first‑derivative test derived from the Mean‑Value Theorem?

f

has a local maximum at

d

✅

f

has a local minimum at

d

f

has an inflection point at

d

D.No inference can be made without further information

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A sign change from positive to negative indicates that the function rises before $d$ and falls after, which characterizes a local maximum. This follows from the first‑derivative test, itself a consequence of the Mean‑Value Theorem. Hence option A is correct.

Q16. Recall the formal statement of the Mean‑Value Theorem.

A.If

f

is continuous on

[a,b]

and differentiable on

(a,b)

, there exists

c\in(a,b)

with

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

. ✅

B.If

f

is continuous on

[a,b]

, then there exists

c\in[a,b]

with

f&#039;(c)=0

C.If

f

is differentiable on

[a,b]

, then it is continuous on

[a,b]

D.If

f

is continuous on

[a,b]

and differentiable on

(a,b)

, then

f

is increasing on

[a,b]

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The Mean‑Value Theorem states that for a function continuous on a closed interval and differentiable on the interior, there exists some interior point where the derivative equals the average rate of change over the interval, i.e., $f'(c)=\frac{f(b)-f(a)}{b-a}$ . This matches option A.

Q17. Given that $f$ is increasing on $[a,b]$ , which of the following must hold for its derivative on $(a,b)$ ?

f&#039;(c)\ge0

for all

c\in(a,b)

✅

f&#039;(c)>0

for all

c\in(a,b)

f&#039;(c)

can be negative at some points

D.None of the above

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If a function is increasing, the Mean‑Value Theorem implies that the average slope over any subinterval is non‑negative, forcing the derivative to be non‑negative at some point in each subinterval. Consequently, the derivative cannot be negative anywhere, so $f'(c)\ge0$ for all interior points, making option A correct.

Q18. Let $f$ be continuous on $[0,2]$ , differentiable on $(0,2)$ with $f(0)=0$ , $f(2)=4$ , and assume $f'(x)\le1$ for all $x$ . Using the Mean‑Value Theorem, can such a function exist?

A.Yes, it can exist

B.No, it cannot exist ✅

C.It exists only if

f

is not differentiable at some point

D.Cannot be determined from the given information

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The average slope over $[0,2]$ is $\frac{4-0}{2}=2$ . The Mean‑Value Theorem guarantees a point where $f'(c)=2$ , contradicting the assumption that $f'(x)\le1$ everywhere. Hence such a function cannot exist, making option B correct.

Q19. Explain how part (c) of Theorem 4.1.2 follows from Rolle’s Theorem, a corollary of the Mean‑Value Theorem.

A.By applying Rolle directly to

f

[a,b]

B.By considering

g(x)=f(x)-k

where

k

is a constant

C.By constructing a function whose derivative is zero ✅

D.By contradiction

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Define $h(x)=f(x)-f(a)$ . Since $h(a)=h(b)=0$ and $h$ inherits continuity and differentiability from $f$ , Rolle’s Theorem provides a point where $h'(c)=0$ . But $h'(c)=f'(c)$ , so $f'(c)=0$ for every interior point, forcing $f$ to be constant. This reasoning matches option C.

Q20. Consider $p(x)=x^{4}-4x^{2}$ . On which intervals is $p$ increasing, according to the derivative test derived from the Mean‑Value Theorem?

(-\infty,-\sqrt{2})\cup(0,\sqrt{2})

(-\infty,-\sqrt{2})\cup(\sqrt{2},\infty)

✅

(-\,\sqrt{2},0)\cup(0,\sqrt{2})

(-\infty,\infty)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The derivative is $p'(x)=4x^{3}-8x=4x(x^{2}-2)$ . This is positive when $x<-\sqrt{2}$ or $x>\sqrt{2}$ . By the Mean‑Value (or first‑derivative) test, the function is increasing exactly on those intervals, which corresponds to option B.

Q21. If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $f(a)=f(b)$ , which theorem guarantees a point $c$ with $f'(c)=0$ ?

A.Rolle’s Theorem ✅

B.Mean‑Value Theorem

C.Intermediate Value Theorem

D.Extreme Value Theorem

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the endpoint values are equal, the hypotheses of Rolle’s Theorem are satisfied, ensuring the existence of an interior point where the derivative vanishes. This is a direct corollary of the Mean‑Value Theorem. Hence option A is correct.

Q22. Define an \increasing function\ as used in Theorem 4.1.2.

f(x_{1})\le f(x_{2})

whenever

x_{1}<x_{2}

f(x_{1})< f(x_{2})

whenever

x_{1}<x_{2}

✅

f&#039;(x)>0

for all

x

in the interval

f

attains its maximum at the right endpoint

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The theorem uses the strict notion of increase: for any two points with $x_{1}<x_{2}$ , the function values satisfy $f(x_{1})<f(x_{2})$ . This excludes equality and matches option B.

Q23. When applying Theorem 4.1.2, which hypothesis would be missing for a function that is continuous on $[a,b]$ but not differentiable on $(a,b)$ ?

A.Continuity on

[a,b]

B.Differentiability on

(a,b)

✅

C.Closedness of the interval

D.Openness of the interval

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The theorem requires the function to be differentiable on the open interval $(a,b)$ . If differenti

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