Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.3 How Derivatives Affect the Shape of a Graph - 3.3 Exercises - Page 229: 33

Answer

a) Increasing on $(-\infty,-2)\cup(2,\infty)$ and $f$, decreasing on $(-2,2)$ b) Local maximum: $f(-2) = 18$, local minimum: $f(2) = -14$ c) Concave upward on $(0,\infty)$, concave downward on $(-\infty,0)$ Inflection point: $(0,2)$ d) See picture below

Work Step by Step

a) $f(x)$ = $x^{3}-12x+2$ => $f'(x)$ = $3x^{2}-12$ = $3(x^{2}-4)$ = $3(x+2)(x-2)$ $f'(x)$ $\gt$ $0$ => $x$ $\lt$ $-2$ or $x$ $\gt$ $2$ $f'(x)$ $\lt$ $0$ => $-2$ $\lt$ $x$ $\lt$ $2$ so $f$ is increasing on $(-\infty,-2)\cup(2,\infty)$ and $f$ is decreasing on $(-2,2)$ b) $f$ changes from increasing to decreasing at $x$ = $-2$ so $f(-2)$ = $18$ is a local maximum value. $f$ changes from decreasing to increasing at $x$ = $2$ so $f(2)$ = $-14$ is a local minimum value c) $f''(x)$ = $6x$ $f''(x)$ = $0\Rightarrow x = 0$ $f''(x)$ $\gt$ $0$ on $(0,\infty)$ $f''(x)$ $\lt$ $0$ on $(-\infty,0)$ so $f$ is concave upward on $(0,\infty)$ and $f$ is concave downward on $(-\infty,0)$ There is an inflection point at $(0,2)$ d) As picture below
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