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