AP Calculus AB Fun 1 C Justify Conclusions About Functions By Applying the Extreme Value Theorem Questions

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 AP Calculus AB Fun 1 C Justify Conclusions About Functions By Applying the Extreme Value Theorem Questions 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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AP Calculus AB Fun 1 C Justify Conclusions About Functions By Applying the Extreme Value Theorem Questions question 1