Question

1. find the critical numbers for $f(x)=sqrt3{4 - x^{2}}$.
2. find the absolute max and min of $h(x)=2x^{3}+3x^{2}-72x + 3$ on $0,5$.
3. find the absolute maximum and minimum of $g(\theta)=sin(\theta)-\theta$ on $-\frac{pi}{2},\frac{pi}{2}$.
4. find the absolute maximum and minimum of $f(x)=5x^{4/5}+4x$ on $-2,1$.
5. find the absolute maximum and minimum of $f(x)=x^{2}+2x$ on $1,2$.

Explanation:

Step1: Recall the definition of critical numbers

Critical numbers of a function $y = f(x)$ are found where $f'(x)=0$ or $f'(x)$ is undefined.
For $f(x)=\sqrt[3]{4 - x^{2}}=(4 - x^{2})^{\frac{1}{3}}$, use the chain - rule. Let $u = 4 - x^{2}$, then $y = u^{\frac{1}{3}}$.
The derivative of $y$ with respect to $u$ is $y'_u=\frac{1}{3}u^{-\frac{2}{3}}$, and the derivative of $u$ with respect to $x$ is $u'_x=-2x$.
By the chain - rule $f'(x)=\frac{1}{3}(4 - x^{2})^{-\frac{2}{3}}\times(-2x)=\frac{-2x}{3\sqrt[3]{(4 - x^{2})^{2}}}$.

Step2: Find where $f'(x) = 0$ and $f'(x)$ is undefined

Set $f'(x)=0$. Then $\frac{-2x}{3\sqrt[3]{(4 - x^{2})^{2}}}=0$. The numerator must be zero for the fraction to be zero, so $-2x = 0$, which gives $x = 0$.
The derivative $f'(x)$ is undefined when the denominator $3\sqrt[3]{(4 - x^{2})^{2}}=0$. But $(4 - x^{2})^{2}\geq0$ for all real $x$, and it is zero when $4 - x^{2}=0$, i.e., $x=\pm2$.

Answer:

The critical numbers of $f(x)=\sqrt[3]{4 - x^{2}}$ are $x=-2,0,2$.