Answer
a) Increasing on $(-1,\infty)$,
decreasing on $(-\infty,-1$
b) Local minimum value: $C(-1) = -3$
c) Concave downward on $(0,2)$,
concave upward on $(-\infty,0)\cup(2,\infty)$
Inflection points: $(0,0)$, $(2,6\sqrt[3] 2)$
d) See graph
Work Step by Step
a)
$C(x)$ = $x^{\frac{1}{3}}(x+4)$ = $x^{\frac{4}{3}}+4x^{\frac{1}{3}}$
$C'(x)$ = $\frac{4}{3}x^{\frac{1}{3}}+\frac{4}{3}x^{-\frac{2}{3}}$ = $\frac{4(x+1)}{3\sqrt[3] {x^{2}}}$
$C'(x)$ $\gt$ $0$ if $-1$ $\lt$ $x$ $\lt$ $0$ or $x$ $\gt$ $0$
$C'(x)$ $\lt$ $0$ if $x$ $\lt$ $-1$
$C$ is increasing on $(-1,\infty)$
$C$ is decreasing on $(-\infty,-1)$
b)
$C(-1)$ = $-3$ is a local minimum value.
c)
$C''(x)$ = $\frac{4}{9}x^{-\frac{2}{3}}-\frac{8}{9}x^{-\frac{5}{3}}(x-2)$ = $\frac{4(x-2)}{9\sqrt[3] {x^{5}}}$
$C''(x)$ $\lt$ $0$ for $0$ $\lt$ $x$ $\lt$ $2$
$C''(x)$ $\gt$ $0$ for $x$ $\lt$ $0$ and $x$ $\gt$ $2$
$C$ is concave downward on $(0,2)$
$C$ is concave upward on $(-\infty,0)\cup(2,\infty)$
There are inflection points at $(0,0)$ and $(2,6\sqrt[3] 2)$.
