Answer
Inflection point : $x=0, x=4$
Concave upward : $(-\infty,0)\cup(4,\infty)$
Concave downward : $(0,4)$
Work Step by Step
The point of inflection is the point where the concavity changes, given by $f''(x)=0$. The graph of $f'(x)$ shows that $f''(x)=0$ at the points on the given graph where the tangent becomes parallel to $ x-$axis. i.e. $x=0$ and $x=4$ are the points of inflection.
The interval in which $f''(x)>0$ is concave upward and $f''(x)<0$ is concave downward. i.e. when $f'(x)$ is increasing the function is concave upward and when $f'(x)$ is decreasing the function is concave downward.
Observe from the graph that $f'(x)$ is increasing in the intervals $(-\infty,0)$ and $(4,\infty)$ while it is decreasing in the intervals $(0,4)$. Hence, the open intervals where the original function is concave upward are $(-\infty,0)\cup(4,\infty)$ and concave downward is $(0,4)$.
