Answer
(a) $h$ is decreasing on the interval $(-2,0)$
$h$ is increasing on the intervals $(-\infty, -2)\cup (0,\infty)$
(b) The local minimum is $h(0) = -1$
The local maximum is $h(-2) = 7$
(c) The graph is concave down on this interval: $(-\infty,-1)$
The graph is concave up on this interval: $(-1, \infty)$
The point of inflection is $(-1, 3)$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $h(x) = (x+1)^5-5x-2$
We can find the points where $h'(x) = 0$:
$h'(x) = 5(x+1)^4-5= 0$
$5(x+1)^4 = 5$
$(x+1)^4 = 1$
$x+1 = \pm(1)^{1/4}$
$x+1 = \pm 1$
$x = -2, 0$
When $-2 \lt x \lt 0~~$ then $h'(x) \lt 0$
$h$ is decreasing on the interval $(-2,0)$
When $x \lt -2$ or $x \gt 0~~$ then $h'(x) \gt 0$
$h$ is increasing on the intervals $(-\infty, -2)\cup (0,\infty)$
(b) $h(-2) = (-2+1)^5-5(-2)-2 = 7$
$h(0) = (0+1)^5-5(0)-2 = -1$
The local minimum is $h(0) = -1$
The local maximum is $h(-2) = 7$
(c) We can find the points where $h''(x) = 0$:
$h''(x) = 20(x+1)^3= 0$
$x = -1$
The graph is concave down when $h''(x) \lt 0$
The graph is concave down on this interval: $(-\infty,-1)$
The graph is concave up when $h''(x) \gt 0$
The graph is concave up on this interval: $(-1, \infty)$
$h(-1) = (-1+1)^5-5(-1)-2 = 3$
The point of inflection is $(-1, 3)$
(d) We can see a sketch of the graph below.