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AP Calculus BC Practice Quiz: Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

Let f be a function that is continuous on the closed interval [2, 10]. According to the Extreme Value Theorem, which of the following is guaranteed?

All Questions (10)

Let f be a function that is continuous on the closed interval [2, 10]. According to the Extreme Value Theorem, which of the following is guaranteed?

A) f has at least one critical point on (2, 10).

B) f has at least one absolute maximum value and one absolute minimum value on [2, 10].

C) f is differentiable on the open interval (2, 10).

D) The absolute maximum value of f must occur at x=10.

Correct Answer: B

The Extreme Value Theorem states that if a function f is continuous over a closed interval [a, b], then it is guaranteed to have at least one minimum value and at least one maximum value on that interval. [cite: 2486]

For which of the following scenarios does the Extreme Value Theorem NOT guarantee the existence of a global maximum and minimum value?

A) A continuous function f on the interval [0, 5].

B) A polynomial function g on the interval [-3, 3].

C) A continuous function h on the interval (0, 5).

D) The function k(x) = |x| on the interval [-1, 1].

Correct Answer: C

The Extreme Value Theorem requires the function to be continuous over a closed interval, such as [a, b]. The interval (0, 5) is an open interval, so the theorem's conditions are not met, and it does not guarantee a global maximum or minimum. [cite: 2486]

A point x=c is defined as a critical point of a function f if it is in the domain of f and which of the following is true?

A) f(c) = 0

B) f has a local extremum at x=c.

C) f'(c) = 0 or f'(c) fails to exist.

D) The graph of f has a horizontal tangent at x=c.

Correct Answer: C

This is the definition of a critical point. A critical point occurs where the first derivative is either equal to zero or is undefined. [cite: 2487]

Which of the following statements correctly describes the relationship between local extrema and critical points?

A) If a function has a critical point at x=c, it must have a local extremum at x=c.

B) If a function has a local extremum at x=c, then x=c must be a critical point.

C) A function can only have local extrema at points where the derivative is zero.

D) Every critical point corresponds to either a local maximum or a local minimum.

Correct Answer: B

According to the provided content, all local (relative) extrema must occur at critical points of a function. However, the converse is not true; not every critical point is a local extremum. [cite: 2488]

What are the critical points of the function g(x) = 2x³ + 3x² - 12x?

A) x = -2 and x = 1

B) x = -1 and x = 2

C) x = 0 only

D) x = -2, x = 0, and x = 1

Correct Answer: A

To find the critical points, we must find where the first derivative equals zero or fails to exist. The derivative is g'(x) = 6x² + 6x - 12. This derivative exists for all x. Setting it to zero: 6(x² + x - 2) = 0, which factors to 6(x+2)(x-1) = 0. The solutions are x = -2 and x = 1. [cite: 2487]

A student is finding the absolute maximum of a continuous function f on a closed interval [a, b]. The procedure involves identifying all critical points in (a, b) and then comparing the function's values at these points and at the endpoints, a and b. Which theorem provides the justification that this procedure will successfully find the absolute maximum?

A) The Mean Value Theorem

B) The Intermediate Value Theorem

C) The Squeeze Theorem

D) The Extreme Value Theorem

Correct Answer: D

The Extreme Value Theorem guarantees that a continuous function on a closed interval will have an absolute maximum and minimum. This theorem is the fundamental justification for the method of checking critical points and endpoints to find these absolute extrema. [cite: 2485, 2486]

The function h(x) = (x-4)^(1/3) has a critical point at x=4. Why is x=4 a critical point?

A) Because h(4) = 0.

B) Because h'(4) = 0.

C) Because h'(4) fails to exist.

D) Because x=4 is an endpoint of the function's domain.

Correct Answer: C

The derivative of the function is h'(x) = (1/3)(x-4)^(-2/3) = 1 / (3(x-4)^(2/3)). The derivative is undefined when the denominator is zero, which occurs at x=4. Since the first derivative fails to exist at x=4, it is a critical point. [cite: 2487]

Let f be a function that is continuous on the closed interval [-1, 5]. If f has a local minimum at x=3, which of the following statements must be true?

A) f'(3) = 0.

B) f(3) is the absolute minimum value of f on [-1, 5].

C) x=3 is a critical point of f.

D) The function f is decreasing for all x < 3.

Correct Answer: C

All local extrema occur at critical points. Therefore, if f has a local minimum at x=3, then x=3 must be a critical point. A critical point is where the derivative is zero OR fails to exist, so we cannot definitively say f'(3)=0. [cite: 2488]

The Extreme Value Theorem guarantees the existence of an absolute maximum and minimum for a function f under what conditions?

A) f is differentiable on a closed interval [a, b].

B) f is continuous on a closed interval [a, b].

C) f is continuous on an open interval (a, b).

D) f has at least one critical point on a closed interval [a, b].

Correct Answer: B

The two required conditions for the Extreme Value Theorem are that the function must be continuous and the interval must be closed. Differentiability is a stronger condition than continuity and is not required. [cite: 2486]

Consider the function f(x) = x³ on the interval [-2, 2]. The function has critical points, and the Extreme Value Theorem applies. Which statement is a correct conclusion?

A) The function has a local extremum at its critical point x=0.

B) The function's absolute maximum occurs at its critical point x=0.

C) The function has a critical point at x=0, but this point is not a local extremum.

D) The function has no critical points on the interval [-2, 2].

Correct Answer: C

The derivative is f'(x) = 3x². Setting f'(x)=0 gives the critical point x=0. However, the function is increasing on both sides of x=0 (since f'(x) is positive for x<0 and x>0), so x=0 is not a local extremum. This illustrates that while all local extrema occur at critical points, not all critical points are local extrema. [cite: 2487, 2488]