Question

consider the function $g(x)=(x - 4)^{\frac{2}{3}}$.
a. find the critical points of $g(x)$. (enter as a comma - separated list, or enter dne for “does not exist” if there are none.)
b. find the absolute extrema of $g$ on the interval $0,12$. enter only the function values.
the absolute maximum is:
the absolute minimum is:

Explanation:

Step1: Differentiate the function

Use the chain - rule. If $y = u^{\frac{2}{3}}$ and $u=x - 4$, then $\frac{dy}{du}=\frac{2}{3}u^{-\frac{1}{3}}$ and $\frac{du}{dx}=1$. So, $g^\prime(x)=\frac{2}{3}(x - 4)^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x - 4}}$.

Step2: Find the critical points

Critical points occur where $g^\prime(x)=0$ or $g^\prime(x)$ is undefined.
Since the numerator of $g^\prime(x)$ is non - zero ($2
eq0$), $g^\prime(x)$ is never $0$. $g^\prime(x)$ is undefined when $x - 4=0$, i.e., $x = 4$. So the critical point is $4$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $g(x)=(x - 4)^{\frac{2}{3}}$ at $x = 0$, $x=4$, and $x = 12$.
When $x = 0$, $g(0)=(0 - 4)^{\frac{2}{3}}=\sqrt[3]{16}$.
When $x = 4$, $g(4)=(4 - 4)^{\frac{2}{3}}=0$.
When $x = 12$, $g(12)=(12 - 4)^{\frac{2}{3}}=\sqrt[3]{64}=4$.

Answer:

a. $4$
b. The absolute maximum is: $4$
The absolute minimum is: $0$