Answer
a. $f'(2)$ does not exist.
b. At $x=2$, $f(x)=0$ minimum.
c. Does not contradict the Extreme Value Theorem.
d. $f'(a)$ does not exist.
At $x=a$, $f(x)=0$ minimum.
Work Step by Step
a. Given the function $f(x)=(x-2)^{2/3}$, we can find its derivative as $f'(x)=\frac{2}{3}(x-2)^{-1/3}=\frac{2}{3\sqrt[3] {x-2}}$. As $f'(x)$ is undefined at $x=2$, $f'(2)$ does not exist.
b. The critical points can be found when $y'=0$ or undefined which happens only when $x=2$. The domain of the function is $(-\infty,\infty)$ and there are no endpoints. At $x=2$, $f(x)=(2-2)^{2/3}=0$ and we can identify this as a minimum (use local test points as necessary) as shown in the figure.
c. The result in part (b) does not contradict the Extreme Value Theorem which states that both a maximum and minimum exist at a closed interval. In this case the domain of the function is $(-\infty,\infty)$ -- the condition of a closed interval is not satisfied.
d. For a function $f(x)=(x-a)^{2/3}$, we can find its derivative as $f'(x)=\frac{2}{3}(x-a)^{-1/3}=\frac{2}{3\sqrt[3] {x-a}}$. As $f'(x)$ is undefined at $x=a$, $f'(a)$ does not exist.
The critical points can be found when $y'=0$ or undefined which happens only when $x=a$. The domain of the function is $(-\infty,\infty)$ and there are no endpoints. At $x=a$, $f(x)=(a-a)^{2/3}=0$ and we can identify this as a minimum ($f(x)\gt0$ everywhere else).
