Answer
a) Increasing on $(-\infty,-2)\cup(0,\infty)$, decreasing on $(-2,0)$
b) Local maximum value: $h(-2) = 7$, local minimum value: $h(0) = -1$
c) Concave upward on $(-1,\infty)$, concave downward on $(-\infty,-1)$
Point of inflection: $(-1,3)$
d) See graph
Work Step by Step
a)
$h(x)$ = $(x+1)^{5}-5x-2$
$h'(x)$ = $5(x+1)^{4}-5$
$h'(x)$ = $0$
$(x+1)^{4}$ = $1$
$x+1$ = $1$ and $x+1$ = $-1$
$x$ = $0$ and $-2$
$h'(x)$ $\gt$ $0\Rightarrow x$ $\lt$ $-2$ or $x$ $\gt$ $0$
$h'(x)$ $\lt$ $0\Rightarrow -2$ $\lt$ $x$ $\lt$ $0$
so
$h$ is increasing on $(-\infty,-2)\cup(0,\infty)$
$h$ is decreasing on $(-2,0)$
b)
$h(-2)$ = $7$ is a local maximum value
$h(0)$ = $-1$ is a local minimum value
c)
$h''(x)$ = $20(x+1)^{3}$ = $0\Rightarrow x$ = $-1$
$h''(x)$ $\gt$ $0\Rightarrow x$ $\gt$ $-1$
$h''(x)$ $\lt$ $0\Rightarrow x$ $\lt$ $-1$
$h$ is concave upward on $(-1,\infty)$
$h$ is concave downward on $(-\infty,-1)$
There is a point of inflection at $(-1,h(-1))$ = $(-1,3)$
