Home / AP Calculus BC: 5.9 Connecting a Function, Its First Derivative, and  Its Second Derivative – Exam Style questions with Answer- FRQ

AP Calculus BC: 5.9 Connecting a Function, Its First Derivative, and  Its Second Derivative – Exam Style questions with Answer- FRQ

Question

The function f is differentiable on the closed interval [−6, 5 ] and satisfies f (−2 )2 = 7. The graph of f ‘, the derivative of f, consists of a semicircle and three line segments, as shown in the figure above.
(a) Find the values of f (−6) and f (5).
(b) On what intervals is f increasing? Justify your answer.
(c) Find the absolute minimum value of f on the closed interval [−6, 5 ]. Justify your answer.
(d) For each of f “(−5) and f “(3), find the value or explain why it does not exist.

Answer/Explanation

Ans:

(a) \(f(-6)=f(-2)+\int_{-2}^{-6}f'(x)dx=7-\int_{-6}^{-2}f'(x)dx=7-4=3\)

\(f(5)=f(-2)+\int_{-2}^{5}f'(x)dx=7-2\pi +3=10-2\pi \)

(b) f ′ ( x) > 0 on the intervals [− 6, -2) and (2, 5). Therefore, f is increasing on the intervals [− 6, -2] and [2, 5] .

(c) The absolute minimum will occur at a critical point where f ‘ (x) = 0 or at an endpoint.
f ′( x) = 0 ⇒x = – 2, x = 2

xf(x)

-6

-2

2

5

3

7

7-2π

10-2π

The absolute minimum value is f (2) = 7 − 2π

(d) \(f”(-5)=\frac{2-0}{-6-(-2)}=-\frac{1}{2}\)

\(\lim_{x\rightarrow 3^{-}}\frac{f'(x)-f'(3)}{x-3}=2 and \lim_{x\rightarrow 3^{+}}\frac{f'(x)-f'(3)}{x-3}=-1\)

f ′′(3) does not exist because

\(\lim_{x\rightarrow 3^{-}}\frac{f'(x)-f'(3)}{x-3}\neq \lim_{x\rightarrow 3^{+}}\frac{f'(x)-f'(3)}{x-3}.\)

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