Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 3 - The Derivative In Graphing And Applications - 3.1 Analysis Of Functions I: Increase, Decrease, and Concavity - Exercises Set 3.1 - Page 195: 32

Answer

See explanation.

Work Step by Step

\[ f^{\prime}(x)=2 \cos 2 x \sin 2 x \cdot 2=2 \sin 4 x \] First derivative Zeros: $x=\pi, x=\frac{3 \pi}{4},x= \pm \frac{\pi}{2}, x=\frac{\pi}{4}$ and $x=0$ (a) Positive on $\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$ , $\left(0, \frac{\pi}{4}\right),$ so the function $f$ is increasing on these intervals (b) Negative on $\left(\frac{3 \pi}{4}, \pi\right)$ , $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, so the function $f$ is decreasing on these intervals. Properties of first derivative \[ 8 \cos 4 x=f^{\prime \prime}(x) \] Second derivative Zeros : $x=\frac{3 \pi}{8},x=\frac{\pi}{8}, x=\frac{7 \pi}{8} $ and $x=\frac{5 \pi}{8}$ (c) Positive on $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$ , $\left[0, \frac{\pi}{8}\right)$ , and $\left(\frac{7 \pi}{8}, \pi\right]$ so the function $f$ is concave up on these intervals. (d) Negative on $\left(\frac{5 \pi}{8}, \frac{7 \pi}{8}\right),$ and $\left(\frac{\pi}{8}, \frac{3 \pi}{8}\right)$ so the function $f$ is concave down on these intervals. Properties of second derivative (e) $x=\frac{7 \pi}{8}, x=\frac{5 \pi}{8}, x=\frac{3 \pi}{8}$ and $,x=\frac{\pi}{8}$ The points where the function changes from concave up to concave down or the other way around are inflection points.
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