AP Calculus AB 5.2 Extreme Value Theorem and Critical Points Question Bank

Practise extrema questions by identifying critical points, comparing local and global behavior, and applying the Extreme Value Theorem.

Exam points

identify critical points and classify local extrema from derivative evidence
use the Extreme Value Theorem to justify global extrema on a closed interval

[Maximum number: 2]

Let f be the function defined above.

Must there be a value of x at which f(x) attains an absolute maximum on the closed interval

$-3 \leq x \leq 4 ?$ Justify your answer.

Figure for Question FUN-1.C—Justify conclusions about functions by applying the Extreme Value Theorem question 1 — AP Calculus AB

Graph of f

$\lim _{x \rightarrow 0^{-}} f(x)=f(0)=3$ and $\lim _{x \rightarrow 0^{+}} f(x)=3$ , so f is continuous at

x=0.

Because f is continuous on [-3, 4], the Extreme Value Theorem

guarantees that f attains an absolute maximum on [-3, 4].

1: answer

$2:\left\{\begin{array}{l}1: f^{\prime}(3) \\ 1: \text { equation }\end{array}\right.$

4: $\left\{\begin{array}{l}1 \text {: integrals of } f \text { over } \\ \quad-3 \leq x \leq 0 \text { and } 0 \leq x \leq 4 \\ 1 \text {: value of } \int_{-3}^{0} \sqrt{9-x^{2}} d x \\ 1 \text {: antiderivative of } \\ \quad-x+3 \cos \left(\frac{\pi x}{2}\right) \\ 1 \text {: answer }\end{array}\right.$ $2:\left\{\begin{array}{l}1: \text { continuity at } x=0 \\ 1: \text { answer with justification }\end{array}\right.$

Question 4

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FUN-1.C—Justify conclusions about functions by applying the Extreme Value Theorem question 1