Answer
a. $f'(0)$ does not exist.
b. $f'(3)$ does not exist.
c. $f'(-3)$ does not exist.
d. At $x=\pm\sqrt 3$, $f(x)=6\sqrt 3$ local maxima.
At $x=0, \pm3$, $f(x)=0$ minima.
Work Step by Step
Given the function $f(x)=|x^3-9x|=\begin{cases} x^3-9x\hspace1cm -3\leq x\leq0, x\geq3 \\-x^3+9x\hspace1cm 0\leq x\leq3, x\leq-3 \end{cases}$, we can find its derivative as $f'(x)=\begin{cases} 3x^2-9\hspace1cm -3\lt x\lt0, x\gt3 \\-3x^2+9\hspace1cm 0\lt x\lt3, x\lt-3 \end{cases}$
a. At $x=0$, $\lim_{x\to0^-}f'(x)=-9$ while $\lim_{x\to0^+}f'(x)=9$, thus $f'(0)$ does not exist.
b. At $x=3$, $\lim_{x\to3^-}f'(x)=-3(3)^2+9=-18$ while $\lim_{x\to3^+}f'(x)=3(3)^2-9=18$, thus $f'(3)$ does not exist.
c. At $x=-3$, $\lim_{x\to-3^-}f'(x)=-3(-3)^2+9=-18$ while $\lim_{x\to-3^+}f'(x)=3(-3)^2-9=18$, thus $f'(-3)$ does not exist.
d. The critical points can be found when $y'=0$ or undefined which happens when $x=\pm\sqrt 3, 0, \pm3$. The domain of the function is $(-\infty,\infty)$ and there are no endpoints.
At $x=\pm\sqrt 3$, $f(x)=|(\sqrt 3)^3-9(\sqrt 3)|=6\sqrt 3$ and we can identify they are local maxima (use local test points as necessary) as shown in the figure.
At $x=0, \pm3$, $f(x)=0$ and we can identify they are minima (use local test points as necessary) as shown in the figure.
