Answer
a. $(0,2), (1,-2), (4,-2)$
b. See graph.
Work Step by Step
a. Step 1. For $f'$ to be defined in the interval of $[-4,6]$, the limit which derives $f'$ needs to exist; more specifically, the left-hand and right-hand limits need to be equal.
Step 2. We can see that at the turning points $(0,2), (1,-2), (4,-2)$, the limits may not exist.
Step 3. At $(0,2)$, $\lim_{x\to0^-}f(x)=\frac{2-0}{0+4}=\frac{1}{2}$ and $\lim_{x\to0^+}f(x)=\frac{-2-2}{1-0}=-4$, thus $f'$ does not exist at this point.
Step 4. At $(1,-2)$, $\lim_{x\to1^-}f(x)=\frac{-2-2}{1-0}=-4$ and $\lim_{x\to1^+}f(x)=\frac{-2+2}{4-1}=0$, thus $f'$ does not exist at this point.
Step 4. At $(4,-2)$, $\lim_{x\to4^-}f(x)=\frac{-2+2}{4-1}=0$ and $\lim_{x\to4^+}f(x)=\frac{2+2}{6-4}=2$ thus $f'$ does not exist at this point.
b. To summarize the results above, we have $f'=\frac{1}{2}$ in $(-4,0)$, $f'=-4$ in $(0,1)$, $f'=0$ in $(1,4)$, and $f'=2$ in $(4,6)$. These results are shown in the figure.
