Answer
(a) $f$ is increasing on the intervals $(-\infty,-2)\cup(2,\infty)$
$f$ is decreasing on the interval $(-2,2)$
(b) The local maximum is $f(-2) = 18$
The local minimum is $f(2) = -14$
(c) The graph is concave down on this interval: $(-\infty,0)$
The graph is concave up on this interval: $(0,\infty)$
The point of inflection is $(0, 2)$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = x^3-12x+2$
We can find the points where $f'(x) = 0$:
$f'(x) = 3x^2-12 = 0$
$3x^2 = 12$
$x^2 = 4$
$x = -2, 2$
When $x \lt -2~~$ or $x \gt 2~~$ then $f'(x) \gt 0$
$f$ is increasing on the intervals $(-\infty,-2)\cup(2,\infty)$
When $-2 \lt x \lt 2$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(-2,2)$
(b) $f(-2) = (-2)^3-12(-2)+2 = 18$
$f(2) = (2)^3-12(2)+2 = -14$
The local maximum is $f(-2) = 18$
The local minimum is $f(2) = -14$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 6x= 0$
$x = 0$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(-\infty,0)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(0,\infty)$
$f(0) = (0)^3-12(0)+2 = 2$
The point of inflection is $(0, 2)$
(d) We can see a sketch of the graph below.