Answer
(a) $C$ is decreasing on the interval $(-\infty,-1)$
$C$ is increasing on the interval $(-1, \infty)$
(b) The local minimum is $C(-1) = -3$
(c) The graph is concave down on this interval: $(0,2)$
The graph is concave up on these intervals: $(-\infty,0)\cup (2, \infty)$
The points of inflection are $(0,0)$ and $(2, 7.56)$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $C(x) = x^{1/3}(x+4) = x^{4/3}+4x^{1/3}$
We can find the points where $C'(x) = 0$:
$C'(x) = \frac{4}{3}x^{1/3}+\frac{4}{3}x^{-2/3}$
$C'(x) = \frac{4}{3}x^{1/3}+\frac{4}{3~x^{2/3}} = 0$
$\frac{4}{3}x^{1/3} = -\frac{4}{3~x^{2/3}}$
$x = -1$
Also note that $C'(x)$ is undefined at $x=0$
When $-\infty \lt x \lt -1~~$ then $C'(x) \lt 0$
$C$ is decreasing on the interval $(-\infty,-1)$
When $-1 \lt x~~$ then $C'(x) \gt 0$
$C$ is increasing on the interval $(-1, \infty)$
(b) $C(-1) = (-1)^{1/3}(-1+4) = -3$
The local minimum is $C(-1) = -3$
There is no local maximum.
(c) We can find the points where $C''(x) = 0$:
$C''(x) = \frac{4}{9}x^{-2/3}-\frac{8}{9~x^{5/3}}$
$C''(x) = \frac{4}{9~x^{2/3}}-\frac{8}{9~x^{5/3}} = 0$
$\frac{4}{9~x^{2/3}}=\frac{8}{9~x^{5/3}}$
$\frac{4x^{5/3}}{9~x^{2/3}}=\frac{8}{9}$
$\frac{4x}{9}=\frac{8}{9}$
$x = 2$
Also note that $C''(x)$ is undefined at $x=0$
The graph is concave down when $G''(x) \lt 0$
The graph is concave down on this interval: $(0,2)$
The graph is concave up when $G''(x) \gt 0$
The graph is concave up on these intervals: $(-\infty,0)\cup (2, \infty)$
$C(0) = (0)^{1/3}(0+4) = 0$
$C(2) = (2)^{1/3}(2+4) = 7.56$
The points of inflection are $(0,0)$ and $(2, 7.56)$
(d) We can see a sketch of the graph below.