Answer
a) $[-2,2]$
b) $(-\infty,-2] $ and $ [2,+\infty)$
c) $(-\infty,0)$
d) $(0,+\infty)$
e) $x=0$
Work Step by Step
$f(x) = 5+12x-x^{3}$
Taking derivatives :
$f'(x) = 12-3x^{2}$
$f''(x) = -6x$
a) Increasing
$f'(x) \gt0$
$12-3x^{2} \gt0$
$3(4-x^{2})\gt0$
$(2-x)(2+x) \gt0$
There are three intervals $(-\infty,-2]$,$[-2,2]$ and $[2,+\infty)$
Checking for $(-\infty,-2]$
Let $x = -3$
$(2-(-3))(2-3) = (2+3)(-1)$
$= -5$ which is not greater than $0$
Checking for $[-2,2]$
Let $x=0$
$(2-0)(2+0) = 4$
As 4 is greater than 0 so it is true
Checking for $[2,+\infty)$
Let $x=3$
$(2-3)(2+3) = (-1)(5)$
$ = -5$ which is not greater than 0
So function is increasing only for $[-2,2]$
b) As we have already calculated in part a) that for $[-2,2]$ function is increasing so for other intervals $(-\infty,-2]$ and $[2,+\infty)$ function will be decreasing.
c) Concave up
$f''(x) \gt0$
$-6x\gt0$
$x\lt0$
Concave up at $(-\infty,0)$
d) Concave down
$f''(x) \lt0$
$-6x\lt0$
$x\gt0$
Concave down at $(0,+\infty)$
e) $f''(x) = 0$
$-6x = 0$
$x = 0$
Since f is concave up at $(-\infty,0)$ and down at $(0,+\infty)$ Hence
$x = 0$ is the inflection point.
