consider the function g(x)=(x - 4)^(2/3). a. find the critical points of g(x). (enter as a comma - separated…

Explanation:

Step1: Find the derivative of (g(x))

First, use the power - rule. If (g(x)=(x - 4)^{\frac{2}{3}}), then by the chain - rule (g^\prime(x)=\frac{2}{3}(x - 4)^{-\frac{1}{3}}\times1=\frac{2}{3(x - 4)^{\frac{1}{3}}}).

Step2: Find the critical points

Critical points occur where (g^\prime(x) = 0) or (g^\prime(x)) is undefined. Set (g^\prime(x)=0), (\frac{2}{3(x - 4)^{\frac{1}{3}}}=0), there is no solution since the numerator is non - zero. (g^\prime(x)) is undefined when (x - 4=0), so (x = 4) is a critical point.

Step3: Evaluate the function at critical points and endpoints

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

Step4: Determine the absolute extrema

Comparing the values (g(0)=\sqrt[3]{16}), (g(4)=0), and (g(12)=4). The absolute minimum value of (g(x)) on ([0,12]) is (0) (at (x = 4)) and the absolute maximum value is (4) (at (x = 12)).

Answer:

a. (4) b. Absolute minimum: (0), Absolute maximum: (4)