9. -/1 points details my notes scalc9 3.1.040...

# Explanation: ## Step1: Recall critical - number definition Critical numbers occur where the derivative is zero or undefined. First, rewrite the function $g(x)=(49 - x^{2})^{\frac{1}{3}}$. ## Step2: Find the derivative using the chain - rule The chain - rule states that if $y = u^{\frac{1}{3}}$ and $u = 49 - x^{2}$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. We know that $\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}$ and $\frac{du}{dx}=-2x$. So, $g^\prime(x)=\frac{-2x}{3(49 - x^{2})^{\frac{2}{3}}}$. ## Step3: Set the derivative equal to zero Set $g^\prime(x) = 0$. Then $\frac{-2x}{3(49 - x^{2})^{\frac{2}{3}}}=0$. The numerator must be zero for the fraction to be zero. So, $-2x = 0$, which gives $x = 0$. ## Step4: Find where the derivative is undefined The derivative $g^\prime(x)$ is undefined when the denominator $3(49 - x^{2})^{\frac{2}{3}}=0$. But $(49 - x^{2})^{\frac{2}{3}} = 0$ when $49 - x^{2}=0$. Solving $49 - x^{2}=0$, we get $x=\pm7$. However, the function $g(x)$ is well - defined at $x = \pm7$. # Answer: $x = - 7,0,7$