Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 302: 48

Answer

(a) $S$ is increasing on the intervals $(0, 2\pi)\cup (2\pi,4\pi)$ (b) There is no local maximum or local minimum. (c) The graph is concave down on these intervals: $(\pi, 2\pi)\cup (3\pi, 4\pi)$ The graph is concave up on these intervals: $(0, \pi)\cup (2\pi, 3\pi)$ The points of inflection are $(\pi,\pi), (2\pi,2\pi)$ and $(3\pi, 3\pi)$ (d) We can see a sketch of the graph below.

Work Step by Step

(a) $S(x) = x- sin~x,~~~~0 \leq x \leq 4\pi$ We can find the points where $S'(x) = 0$: $S'(x) = 1 -cos ~x= 0$ $cos~x = 1$ $x = 0,2\pi, 4\pi$ Since $S'(x) \geq 0$ for all $x$, $S(x)$ is not decreasing at any point. When $0 \lt x \lt 2\pi$ or $2\pi \lt x \lt 4\pi~~$ then $S'(x) \gt 0$ $S$ is increasing on the intervals $(0, 2\pi)\cup (2\pi,4\pi)$ (b) There is no local maximum or local minimum, because $S'(x) \geq 0$ for all $x$ (c) We can find the points where $S''(x) = 0$: $S''(x) = sin~x = 0$ $x = 0, \pi, 2\pi, 3\pi, 4\pi$ The graph is concave down when $S''(x) \lt 0$ The graph is concave down on these intervals: $(\pi, 2\pi)\cup (3\pi, 4\pi)$ The graph is concave up when $S''(x) \gt 0$ The graph is concave up on these intervals: $(0, \pi)\cup (2\pi, 3\pi)$ $S(\pi) = \pi- sin~\pi = \pi$ $S(2\pi) = 2\pi- sin~2\pi = 2\pi$ $S(3\pi) = 3\pi- sin~3\pi = 3\pi$ The points of inflection are $(\pi,\pi), (2\pi,2\pi)$ and $(3\pi, 3\pi)$ (d) We can see a sketch of the graph below.
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