Answer
(a) $f$ is decreasing on the intervals $(-\infty,-2)\cup(3,\infty)$
$f$ is increasing on the interval $(-2,3)$
(b) The local maximum is $f(3) = 81$
The local minimum is $f(-2) = -44$
(c) The graph is concave down on this interval: $(\frac{1}{2},\infty)$
The graph is concave up on this interval: $(-\infty,\frac{1}{2})$
The point of inflection is $(\frac{1}{2}, \frac{37}{2})$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = 36x+3x^2-2x^3$
We can find the points where $f'(x) = 0$:
$f'(x) = 36+6x-6x^2 = 0$
$x^2-x-6 = 0$
$(x-3)(x+2)=0$
$x = -2,3$
When $x \lt -2~~$ or $x \gt 3~~$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(-\infty,-2)\cup(3,\infty)$
When $-2 \lt x \lt 3$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(-2,3)$
(b) $f(-2) = 36(-2)+3(-2)^2-2(-2)^3 = -44$
$f(3) = 36(3)+3(3)^2-2(3)^3 = 81$
The local maximum is $f(3) = 81$
The local minimum is $f(-2) = -44$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 6-12x= 0$
$2x-1 = 0$
$x = \frac{1}{2}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(\frac{1}{2},\infty)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-\infty,\frac{1}{2})$
$f(\frac{1}{2}) = 36(\frac{1}{2})+3(\frac{1}{2})^2-2(\frac{1}{2})^3 = \frac{37}{2}$
The point of inflection is $(\frac{1}{2}, \frac{37}{2})$
(d) We can see a sketch of the graph below.