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📝 Absolute extrema with one critical point (23 MCQs)

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

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What is Absolute extrema with one critical point?

Definition:
If a continuous function on an interval has only one critical point, that point is often the absolute extremum if the function behaves monotonically elsewhere. Verification involves checking endpoint values or limits to confirm if the critical point represents the global peak or valley.

Example:
For $f(x) = -x^2$ on $(-\infty, \infty)$ , the single critical point $x=0$ is the absolute maximum since $f(x) < 0$ for all $x \ne 0$ .

Reason:
Simplifies optimization by focusing on the unique turning point, assuming end behavior does not exceed this value, which is common in quadratic and simple polynomial models.

Easy

Medium

Hard

📝 All Absolute extrema with one critical point MCQs

Q1. A continuous function f on (0,∞) has exactly one relative maximum at x=2. Which of the following must be true?

A.f attains its absolute maximum at x=2 ✅

B.f attains its absolute minimum at x=2

C.f has no absolute extremum on (0,∞)

D.Both absolute maximum and minimum occur at x=2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By the theorem, if a continuous function has only one relative extremum, that extremum is also absolute. Since the unique relative extremum is a maximum at x=2, it must be the absolute maximum on the interval, guaranteeing option A is correct.

Q2. Suppose f is continuous on [−5,5] and has a single relative minimum at x=−1. Which conclusion follows regarding absolute extrema?

A.f has an absolute minimum at x=−1 ✅

B.f has an absolute maximum at x=−1

C.f has both absolute maximum and minimum at x=−1

D.No absolute extremum exists on [−5,5]

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The theorem asserts that the sole relative extremum becomes the absolute extremum of the same type. Hence the lone relative minimum at x=−1 is also the absolute minimum on the closed interval, making option A the correct inference.

Q3. If a function g(x)=e^{x^3-3x^2} is defined on (0,∞) and its derivative vanishes only at x=2, what can be deduced about its absolute extrema?

A.g has an absolute minimum at x=2 ✅

B.g has an absolute maximum at x=2

C.g has no absolute extrema on (0,∞)

D.g has both absolute maximum and minimum at x=2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative of g is zero only at x=2, giving a single relative extremum. Since g is continuous and the only extremum is a relative minimum (second‑derivative test positive), it must also be the absolute minimum on the interval, confirming option A.

Q4. Consider a continuous function h on ℝ with exactly one relative extremum, a maximum at x=0. Which scenario is impossible?

A.h attains its absolute maximum at x=0

B.h attains its absolute minimum at x=0 ✅

C.h is unbounded above

D.h has no critical points other than x=0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If the sole relative extremum is a maximum, the function cannot simultaneously have an absolute minimum at the same point because the theorem guarantees the extremum’s type (maximum) carries over to the absolute case. Therefore, option B contradicts the theorem and is impossible.

Q5. A function p(x) is continuous on [−2,4] and has exactly one critical point at x=1 where p'(1)=0. Which statement correctly describes the absolute extrema?

A.p has an absolute extremum at x=1, but its type cannot be determined without further info ✅

B.p has an absolute maximum at x=1

C.p has an absolute minimum at x=1

D.p has neither absolute maximum nor minimum on [−2,4]

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: With only one critical point, the relative extremum at x=1 must be either a maximum or a minimum. However, without the second‑derivative sign or endpoint values, we cannot decide which type, so only the existence of an absolute extremum at x=1 is guaranteed.

Q6. For the function q(x)=x^3-3x^2+2 on (−∞,∞), the derivative q'(x)=3x^2-6x has two zeros. Which conclusion about absolute extrema is correct?

A.q cannot have an absolute extremum because it has more than one relative extremum ✅

B.q has an absolute maximum at the larger critical point

C.q has an absolute minimum at the smaller critical point

D.The theorem does not apply, so no conclusion can be drawn

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The theorem requires exactly one relative extremum. Since q'(x) has two distinct zeros, q possesses two relative extrema, violating the hypothesis. Hence we cannot assert that any relative extremum is absolute, making option A the appropriate conclusion.

Q7. Let r(x) be continuous on (0,∞) and satisfy r'(x)>0 for all x>0 except at x=3 where r'(3)=0. What does the theorem imply about r?

A.r has an absolute maximum at x=3

B.r has an absolute minimum at x=3 ✅

C.r is strictly increasing and has no absolute extremum

D.r has both absolute maximum and minimum at x=3

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since r'(x) is positive everywhere except a single zero at x=3, the function has one relative extremum, which must be a minimum because the derivative changes from positive to positive (flat). By the theorem, this relative minimum is also absolute, so option B is correct.

Q8. If a continuous function s(x) on [0,10] has exactly one relative extremum, a minimum at x=7, and s(0)=s(10)=5, what can be concluded about its absolute maximum?

A.The absolute maximum occurs at either endpoint 0 or 10 ✅

B.The absolute maximum occurs at x=7

C.The absolute maximum does not exist

D.The absolute maximum is less than s(7)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: With only one relative extremum (a minimum), the theorem tells us that the minimum at x=7 is also absolute. The maximum must then lie at the interval’s endpoints, where the function values are 5, confirming that the absolute maximum occurs at either x=0 or x=10.

Q9. A function f(x)=e^{x^3-3x^2} is defined on (0,∞). Which of the following best explains why f lacks an absolute maximum on this interval?

A.Because f(x)→∞ as x→∞, so it cannot attain a greatest value ✅

B.Because the derivative never vanishes

C.Because the function is not continuous at 0

D.Because the relative minimum prevents a maximum

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The limit $\displaystyle\lim_{x\to +\infty}e^{x^3-3x^2}=+\infty$ ; thus the function grows without bound and never reaches a greatest finite value, precluding an absolute maximum, which validates option A.

Q10. Consider a function t(x) continuous on (−∞,∞) that has exactly one relative minimum at x=−2. Which statement about the behavior of t(x) as x→±∞ must be true?

A.t(x)→+∞ as x→±∞

B.t(x)→−∞ as x→±∞

C.t(x) is bounded above but not below

D.No specific behavior can be inferred from the given information ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The theorem only links the unique relative extremum to an absolute extremum of the same type; it imposes no restrictions on the function’s end behavior. Therefore, we cannot deduce whether t(x) tends to plus or minus infinity, making option D correct.

Q11. Which of the following functions satisfies the hypothesis of the theorem (continuous with exactly one relative extremum) and has its absolute extremum at x=0?

A.f(x)=x^4

B.f(x)=\sin x

C.f(x)=e^{-x^2} ✅

D.f(x)=\ln(x+1)

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The function $f(x)=e^{-x^2}$ is continuous everywhere and has a single relative maximum at x=0 (since its derivative vanishes only there and the second derivative is negative). By the theorem, this relative maximum is also absolute, confirming option C.

Q12. A continuous function u on [−3,3] has exactly one critical point at x=0 where u'(0)=0 and u''(0)>0. Which conclusion follows?

A.u has an absolute minimum at x=0 ✅

B.u has an absolute maximum at x=0

C.u has both absolute maximum and minimum at x=0

D.No absolute extremum can be guaranteed

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The positive second derivative indicates a relative minimum at x=0. Since there is only one relative extremum, the theorem guarantees that this relative minimum is also the absolute minimum on the closed interval, making option A correct.

Q13. If a function v(x) is continuous on (−1,1) and its only critical point is at x=0 where v'(0)=0, but v''(0)=0, what can be inferred about absolute extrema?

A.v must have an absolute extremum at x=0

B.v cannot have any absolute extremum on (−1,1)

C.The theorem does not apply because the second derivative test is inconclusive ✅

D.v has both absolute maximum and minimum at x=0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The theorem requires only the existence of a single relative extremum; however, with v''(0)=0, we cannot determine whether x=0 is a relative extremum without further analysis. Consequently, the theorem’s hypothesis is not satisfied, and we cannot conclude any absolute extremum, so option C is appropriate.

Q14. Which logical implication correctly describes the relationship between a single relative extremum and absolute extremum for a continuous function on a closed interval?

A.If a function has one relative extremum, then it must be both absolute maximum and minimum

B.If a function has one relative extremum, then that extremum is also absolute of the same type ✅

C.If a function has one absolute extremum, then it must have exactly one relative extremum

D.If a function has one relative extremum, then it has no endpoints affecting extrema

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The theorem states that for a continuous function with exactly one relative extremum, that extremum is also absolute and retains its nature (maximum or minimum). Therefore, the correct logical implication is option B.

Q15. Compare the behavior of a function w(x) that satisfies the theorem on an infinite interval versus a finite closed interval. Which statement is accurate?

A.On an infinite interval, the function cannot have an absolute extremum

B.On a finite interval, the absolute extremum must occur at an endpoint

C.The type of absolute extremum (max or min) is independent of interval finiteness ✅

D.The theorem only applies to finite intervals

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The theorem applies to both finite and infinite intervals; it guarantees that the unique relative extremum is also absolute, regardless of whether the interval is bounded. Hence, the type of absolute extremum is independent of interval finiteness, making option C correct.

Q16. A function y(x) = e^{x^3-3x^2} is defined on (0,∞). Which analytical step confirms that x=2 yields an absolute minimum?

A.Evaluating

\displaystyle\lim_{x\to0^+}y(x)=1

and

\displaystyle\lim_{x\to\infty}y(x)=\infty

B.Checking that y'(2)=0 and y''(2)>0

C.Showing that y is decreasing for x<2 and increasing for x>2

D.All of the above ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: To verify that x=2 gives an absolute minimum, we need to (1) locate the critical point by y'(2)=0, (2) confirm it is a relative minimum via y''(2)>0, and (3) examine the limits at the interval’s ends to ensure no lower values exist. All three steps are required, so option D is correct.

Q17. Which of the following statements is a direct consequence of the theorem for a function defined on (−∞,∞) with exactly one relative maximum?

A.The function must be bounded above

B.The function must attain its maximum at a finite x-value ✅

C.The function cannot have any asymptotes

D.The function is periodic

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The theorem guarantees that the single relative maximum is also the absolute maximum, meaning the function attains its greatest value at that specific finite x-coordinate. It does not require boundedness overall or any other property, so option B follows directly.

Q18. In the example where f(x)=e^{x^3-3x^2} on (0,∞), why is there no absolute maximum?

A.Because f'(x) never changes sign

B.Because f(x) approaches infinity as x→∞ ✅

C.Because the only critical point is a minimum

D.Because the function is not defined at x=0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The limit $\displaystyle\lim_{x\to+\infty}e^{x^3-3x^2}=+\infty$ shows the function grows without bound, so it cannot achieve a greatest finite value; thus, no absolute maximum exists, confirming option B.

Q19. A student claims that if a function has exactly one relative extremum, then the function must be monotonic elsewhere. Is this claim always true?

A.Yes, because the theorem forces monotonicity away from the extremum

B.No, the function may oscillate without creating additional extrema ✅

C.Yes, otherwise a second extremum would appear

D.The claim is unrelated to continuity

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Having a single relative extremum does not preclude the function from oscillating in a way that never creates another extremum (e.g., flattening). Therefore, monotonicity is not guaranteed, making the student's claim false; option B correctly reflects this.

Q20. Which conceptual principle explains why a continuous function with one relative extremum cannot have a higher value elsewhere without creating another extremum?

A.Intermediate Value Theorem

B.Mean Value Theorem

C.Extreme Value Theorem

D.Rolle’s Theorem ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Rolle’s Theorem states that if a function attains equal values at two points, there must be a point where the derivative is zero between them. To achieve a higher value without a new extremum would violate this principle, so the conceptual justification relies on Rolle’s Theorem, option D.

Q21. Given a function p(x) that is continuous on [a,b] and has exactly one relative maximum at c∈(a,b), which of the following must be true about p(a) and p(b)?

A.Both must be less than p(c)

B.Both must be greater than p(c)

C.At least one of them must be less than p(c) ✅

D.No relationship can be deduced

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since c is the absolute maximum by the theorem, any value at the endpoints cannot exceed p(c). However, one or both endpoints may be equal or lower; the only guaranteed statement is that at least one endpoint value is not greater than p(c), making option C correct.

Q22. What is the definition of a relative extremum for a continuous function?

A.A point where the function equals its global maximum or minimum

B.A point where the function’s derivative does not exist

C.A point where the function attains a local maximum or minimum within a neighborhood ✅

D.A point where the function is continuous

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A relative extremum (also called a local extremum) occurs at a point where the function achieves a maximum or minimum relative to nearby points, i.e., within some open interval around that point. This matches option C.

Q23. Which of the following best describes the role of the second derivative test in applying the theorem?

A.It determines continuity of the function

B.It confirms the type (max or min) of the single relative extremum ✅

C.It locates all critical points

D.It provides the exact value of the absolute extremum

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When a function has a single critical point, the second derivative test helps identify whether that point is a relative maximum or minimum. Knowing the type allows the theorem to assert that the same point is the absolute extremum of that type, so option B is correct.

🔗 Related Topics (MCQs)

📝 Increasing and decreasing functions using derivative MCQs 📝 Concavity and inflection points calculus MCQs 📝 Logistic growth curves calculus MCQs 📝 Analysis of Functions II (Relative Extrema) in calculus MCQs 📝 Relative maxima and minima local extrema MCQs

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