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: 29

Answer

See explanation.

Work Step by Step

\[ -\frac{1}{2} \sec ^{2} \frac{x}{2}=f^{\prime}(x) \] First derivative Zeros: None (because sec $x$ does not have roots) (a) Never positive, so the function $f$ is never increasing. (b) Negative everywhere, thus the function $f$ is decreasing everywhere $(\text { on }(-\pi, \pi))$ $ f^{\prime \prime}(x)=-\sec \frac{x}{2} \cdot \sec \frac{x}{2} \tan \frac{x}{2} \cdot \frac{1}{2}$ Zeros : $0=x$ (c) Positive on $(-\pi, 0)$ so the function $f$ is concave up on this interval. (d) Negative on $(0, \pi),$ so the function $f$ is concave down on this interval. Properties of second derivative (e) $0=x$ The points where the function changes from concave up to concave down or the other way around are inflection points.
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