Answer
a)
f increasing on $(-\infty,-1)$ $∪$ $(3,\infty)$
f decreasing on $(-1,3)$
b)
$f(−1) = 9$ is a local maximum value and
$f(3) = −23$ is a local minimum value.
c)
$f$ is concave upward on $(1,\infty)$
$f$ is concave downward on $(-\infty,1)$
Inflection point at $(1,-7)$
Work Step by Step
a)
$f(x) = x^{3}-3x^{2}-9x+4$
$f'(x) = 3x^{2}-6x-9 = 3(x-3)(x+1)$
$f$ increasing on $(-\infty,-1)$ $∪$ $(3,\infty)$
$f$ decreasing on $(-1,3)$
b)
$f$ changes from increasing to decreasing at $x = −1$ and from decreasing to increasing at $x = 3$. Thus,
$f(−1) = 9$ is a local maximum value and
$f(3) = −23$ is a local minimum value.
c)
$f''(x) = 6x-6 = 6(x-1)$
$f''(x)\gt 0$ for $x\gt 1$
$f''(x)\lt 0$ for $x\lt 1$
Thus
$f$ is concave upward on $(1,\infty)$
$f$ is concave downward on $(-\infty,1)$
There is an inflection point at $(1,-7)$
