Question

consider the function g(x)=(x - 4)^(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: 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: \(4\)