Answer
a) Increasing on $(-\infty,-2)\cup(2,\infty)$ and $f$, decreasing on $(-2,2)$
b) Local maximum: $f(-2) = 18$, local minimum: $f(2) = -14$
c) Concave upward on $(0,\infty)$, concave downward on $(-\infty,0)$
Inflection point: $(0,2)$
d) See picture below
Work Step by Step
a)
$f(x)$ = $x^{3}-12x+2$ => $f'(x)$ = $3x^{2}-12$ = $3(x^{2}-4)$ = $3(x+2)(x-2)$
$f'(x)$ $\gt$ $0$ => $x$ $\lt$ $-2$ or $x$ $\gt$ $2$
$f'(x)$ $\lt$ $0$ => $-2$ $\lt$ $x$ $\lt$ $2$
so
$f$ is increasing on $(-\infty,-2)\cup(2,\infty)$ and $f$ is decreasing on $(-2,2)$
b)
$f$ changes from increasing to decreasing at $x$ = $-2$ so $f(-2)$ = $18$ is a local maximum value. $f$ changes from decreasing to increasing at $x$ = $2$ so $f(2)$ = $-14$ is a local minimum value
c)
$f''(x)$ = $6x$
$f''(x)$ = $0\Rightarrow x = 0$
$f''(x)$ $\gt$ $0$ on $(0,\infty)$
$f''(x)$ $\lt$ $0$ on $(-\infty,0)$
so $f$ is concave upward on $(0,\infty)$ and $f$ is concave downward on $(-\infty,0)$
There is an inflection point at $(0,2)$
d) As picture below