Science:Math Exam Resources/Courses/MATH104/December 2016/Question 08 (a)/Solution 1

From the graph of $y=f'(x)$, we can first see that $f'(2)=f'(4)=0$, so that both are critical points; the candidates of the local extrema.

Since $f'>0$ on $(0,2)\cup (4,5)$, while $f'<0$ on $(2,4)$, $f$ is increasing on $(0,2)\cup (4,5)$ and decreasing on $(2,4)$. This implies that $x=2$ is a local maximum and $x=4$ is a local minimum.

On the other hand, we observe that $f''(1)=f''(3)=0$, and hence both are the candidates of inflection points. Indeed, this follows from that $x=1$ and $x=3$ are local extrema of $f'$ and hence the critical points of $f'$.

Finally, the graph tells us that $f'$ increases on the interval $(0,1)$ and $(3,5)$, while it decreases on $(1,3)$. This determines the sign of the second derivative of $f$ and hence the concavity of the function $f$ as follows;

Intervals	$(0,1)$	$(1,3)$	$(3,5)$
Sign of $f''$	$+$	$-$	$+$
Concavity	Up	Down	Up

Since the sign of the second derivative is changing at both points $x=1$ and $x=3$, they are inflection points of $f$.

To summarize, we obtain a local maximum $x=2$ and a local minimum $x=4$ and two inflection points $x=1,3$. Then, based on our observation, the graph of $f$ can be sketched as follows;