Answer
(a) $g$ is decreasing on the interval $(-\infty,-6)$
$g$ is increasing on the interval $(-6,\infty)$
(b) The local minimum is $g(-6) = -232$
There is no local maximum.
(c) The graph is concave down on this interval: $(-4,0)$
The graph is concave up on these intervals: $(-\infty,-4)\cup (0, \infty)$
The points of inflection are $(-4, -56)$ and $(0, 200)$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $g(x) = 200+8x^3+x^4$
We can find the points where $g'(x) = 0$:
$g'(x) = 24x^2+4x^3 = 0$
$4x^2(6+x) = 0$
$x = 0, -6$
When $x \lt -6~~$ then $g'(x) \lt 0$
$g$ is decreasing on the interval $(-\infty,-6)$
When $-6 \lt x~~$ then $g'(x) \gt 0$
$g$ is increasing on the interval $(-6,\infty)$
(b) $g(-6) = 200+8(-6)^3+(-6)^4 = -232$
The local minimum is $g(-6) = -232$
There is no local maximum.
(c) We can find the points where $g''(x) = 0$:
$g''(x) = 48x+12x^2= 0$
$12x(4+x)= 0$
$x = -4, 0$
The graph is concave down when $g''(x) \lt 0$
The graph is concave down on this interval: $(-4,0)$
The graph is concave up when $g''(x) \gt 0$
The graph is concave up on these intervals: $(-\infty,-4)\cup (0, \infty)$
$g(-4) = 200+8(-4)^3+(-4)^4 = -56$
$g(0) = 200+8(0)^3+(0)^4 = 200$
The points of inflection are $(-4, -56)$ and $(0, 200)$
(d) We can see a sketch of the graph below.