Question

1. find the critical numbers for (f(x)=sqrt3{4 - x^{2}}).

Explanation:

Step1: Recall critical - number definition

Critical numbers of a function \(y = f(x)\) are found where \(f^{\prime}(x)=0\) or \(f^{\prime}(x)\) is undefined. First, find the derivative of \(y = f(x)=\sqrt[3]{4 - x^{2}}=(4 - x^{2})^{\frac{1}{3}}\) using the chain - rule. The chain - rule states that if \(y = u^{\frac{1}{3}}\) and \(u = 4 - x^{2}\), then \(y^{\prime}=\frac{dy}{du}\cdot\frac{du}{dx}\).
\(\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}\) and \(\frac{du}{dx}=-2x\).
So \(f^{\prime}(x)=\frac{1}{3}(4 - x^{2})^{-\frac{2}{3}}\cdot(-2x)=\frac{-2x}{3\sqrt[3]{(4 - x^{2})^{2}}}\).

Step2: Find where \(f^{\prime}(x) = 0\)

Set \(f^{\prime}(x)=0\).
\(\frac{-2x}{3\sqrt[3]{(4 - x^{2})^{2}}}=0\).
Since the denominator \(3\sqrt[3]{(4 - x^{2})^{2}}
eq0\) for all real \(x\) (the cube - root of a non - negative number \((4 - x^{2})^{2}\) is non - zero except when \(4 - x^{2}=0\), but we are looking at the fraction being zero), we set the numerator equal to zero.
\(-2x = 0\), which gives \(x = 0\).

Step3: Find where \(f^{\prime}(x)\) is undefined

The function \(f^{\prime}(x)=\frac{-2x}{3\sqrt[3]{(4 - x^{2})^{2}}}\) is undefined when the denominator \(3\sqrt[3]{(4 - x^{2})^{2}} = 0\).
Set \(4 - x^{2}=0\).
\(x^{2}=4\), so \(x=\pm2\).

Answer:

\(x=-2,0,2\)