Answer
The table in Exercise 12 matches with graph $(b)$.
Work Step by Step
- Here we see that at $a$ and $b$, we have $f'(a)=f'(b)=0$. According to Theorem 2, this means there are local maximum or minimum values at $x=a$ and $x=b$.
Looking at the graphs, we find that graphs $(b)$ and $(c)$ satisfy this fact.
In graph $(a)$, the sharp points at $x=a$ and $x=b$ indicate that $f'(x)$ is undefined at these places.
In graph $(d)$, again, the sharp point at $x=a$ indicates that $f'(x)$ is undefined at $x=a$.
- We also have $f'(c)=-5\lt0$. $f'(c)$ being negative means that the graph at $x=c$ must be falling. Graph $(b)$ satisfies this fact.
Therefore, the table in Exercise 12 matches with graph $(b)$.
