AP Calculus AB 5.2 Extrema & Critical Points Review

Review AP Calculus AB 5.2 by locating critical points, comparing local and absolute extrema, and checking the Extreme Value Theorem on closed intervals.

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

Covered syllabus points

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

[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 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points 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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Covered syllabus points

5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points question 1