Answer
(a) $f$ is increasing on these intervals: $(0,2)\cup (4,6)\cup (8,\infty)$
$f$ is decreasing on these intervals: $(2,4)\cup (6,8)$
(b) $f$ has a local maximum at $x=2$
$f$ has a local minimum at $x=4$ and $x=8$
(c) $f$ is concave up on these intervals: $(3,6)\cup (6,9)$
$f$ is concave down on this interval: $(0,3)$
(d) The x-coordinate of the point of inflection is $x=3$
(e) We can see a sketch of the graph of $f$ below.
Work Step by Step
(a) $f$ is increasing when $f'(x) \gt 0$
$f$ is increasing on these intervals: $(0,2)\cup (4,6)\cup (8,\infty)$
$f$ is decreasing when $f'(x) \lt 0$
$f$ is decreasing on these intervals: $(2,4)\cup (6,8)$
(b) $f$ has a local maximum when $f'(x)$ changes from positive to negative.
$f$ has a local maximum at $x=2$.
Satisfying condition C in the First Derivative Test, If $f'$ is positive to the left and right of $c$, or negative to the left and right of $c$, then $f$ has no local maximum or minimum at $c$. Therefore, $x=6$ cannot be a local maximum or minimum.
$f$ has a local minimum when $f'(x)$ changes from negative to positive.
$f$ has a local minimum at $x=4$ and $x=8$
(c) $f$ is concave up when $f'(x)$ is increasing.
$f$ is concave up on these intervals: $(3,6)\cup (6,\infty)$
$f$ is concave down when $f'(x)$ is decreasing.
$f$ is concave down on this interval: $(0,3)$
(d) A point of inflection is a point where the graph $f$ changes concavity.
$f$ changes concavity at $x = 3$ so the x-coordinate of the point of inflection is $x=3$
(e) We can see a sketch of the graph of $f$ below.
