Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises - Page 267: 64

Answer

(a) $f'(x) = 4-sec^2~x$ $f''(x) = -2~sec^2~x~tan~x$ (b) We can see a sketch of the graphs below.
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Work Step by Step

(a) $-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$ $f(x) = 4x-tan~x$ We can find $f'(x)$: $f'(x) = 4-sec^2~x$ We can find $f''(x)$: $f''(x) = (-2~sec~x)~(sec~x~tan~x)$ $f''(x) = -2~sec^2~x~tan~x$ (b) We can see a sketch of the graphs below. $f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope. $f''(x)$ seems reasonable since $f''(x)$ has negative values when the slope of $f'(x)$ has a negative slope, $f''(x)$ is 0 when the slope of $f'(x)$ is 0, and $f''(x)$ has positive values when the slope of $f'(x)$ has a positive slope.
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