Answer
(a) $h$ is decreasing on the interval $(-\infty,-1)\cup (1, \infty)$
$h$ is increasing on the intervals $(-1, 0)\cup (0,1)$
(b) The local minimum is $h(-1) = -2$
The local maximum is $h(1) = 2$
(c) The graph is concave down on these intervals: $(-\frac{\sqrt{2}}{2},0)\cup (\frac{\sqrt{2}}{2}, \infty)$
The graph is concave up on these intervals: $(-\infty, -\frac{\sqrt{2}}{2})\cup (0, \frac{\sqrt{2}}{2})$
The points of inflection are $(-\frac{\sqrt{2}}{2},-\frac{7\sqrt{2}}{8}), (0, 0),$ and $(\frac{\sqrt{2}}{2},\frac{7\sqrt{2}}{8})$
Work Step by Step
(a) $h(x) = 5x^3-3x^5$
We can find the points where $h'(x) = 0$:
$h'(x) = 15x^2-15x^4= 0$
$ 15x^2(1-x^2)= 0$
$x = -1, 0,1$
When $x \lt -1~~$ or $x \gt 1$ then $h'(x) \lt 0$
$h$ is decreasing on the interval $(-\infty,-1)\cup (1, \infty)$
When $-1 \lt x \lt 0$ or $0 \lt x \lt 1~~$ then $h'(x) \gt 0$
$h$ is increasing on the intervals $(-1, 0)\cup (0,1)$
(b) $h(-1) = 5(-1)^3-3(-1)^5 = -2$
$h(0) = 5(0)^3-3(0)^5 = 0$
$h(1) = 5(1)^3-3(1)^5 = 2$
The local minimum is $h(-1) = -2$
The local maximum is $h(1) = 2$
(c) We can find the points where $h''(x) = 0$:
$h''(x) = 30x-60x^3= 0$
$30x(1-2x^2) = 0$
$x = -\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}$
The graph is concave down when $h''(x) \lt 0$
The graph is concave down on these intervals: $(-\frac{\sqrt{2}}{2},0)\cup (\frac{\sqrt{2}}{2}, \infty)$
The graph is concave up when $h''(x) \gt 0$
The graph is concave up on these intervals: $(-\infty, -\frac{\sqrt{2}}{2})\cup (0, \frac{\sqrt{2}}{2})$
$h(-\frac{\sqrt{2}}{2}) = 5(-\frac{\sqrt{2}}{2})^3-3(-\frac{\sqrt{2}}{2})^5 = -\frac{7\sqrt{2}}{8}$
$h(0) = 5(0)^3-3(0)^5 = 0$
$h(\frac{\sqrt{2}}{2}) = 5(\frac{\sqrt{2}}{2})^3-3(\frac{\sqrt{2}}{2})^5 = \frac{7\sqrt{2}}{8}$
The points of inflection are $(-\frac{\sqrt{2}}{2},-\frac{7\sqrt{2}}{8}), (0, 0),$ and $(\frac{\sqrt{2}}{2},\frac{7\sqrt{2}}{8})$
(d) We can see a sketch of the graph below.