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.
