9. find the critical numbers for $f(x)=sqrt3{4 - x^{2}}$.\n10. find the absolute max and min of…

Explanation:

Step1: Recall the definition of critical numbers

Critical numbers of a function $y = f(x)$ are the values of $x$ in the domain of $f$ where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ does not exist.

Step2: Differentiate $f(x)=\sqrt[3]{4 - x^{2}}$

Rewrite $f(x)=(4 - x^{2})^{\frac{1}{3}}$. Using the chain - rule, if $u = 4 - x^{2}$, then $y = u^{\frac{1}{3}}$. The derivative of $y$ with respect to $u$ is $y^{\prime}_u=\frac{1}{3}u^{-\frac{2}{3}}$ and the derivative of $u$ with respect to $x$ is $u^{\prime}_x=-2x$. So, $f^{\prime}(x)=\frac{-2x}{3(4 - x^{2})^{\frac{2}{3}}}$.

Step3: Find where $f^{\prime}(x) = 0$

Set $f^{\prime}(x)=0$. Then $\frac{-2x}{3(4 - x^{2})^{\frac{2}{3}}}=0$. Since the denominator $3(4 - x^{2})^{\frac{2}{3}}\neq0$ for all real $x$ (the denominator is non - zero because the cube - root of a non - zero number is non - zero), we solve $-2x = 0$, which gives $x = 0$.

Step4: Find where $f^{\prime}(x)$ does not exist

The derivative $f^{\prime}(x)$ does not exist when the denominator $3(4 - x^{2})^{\frac{2}{3}}=0$. Solve $4 - x^{2}=0$, which gives $x=\pm2$.

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

Answer:

$x=-2,0,2$

Explanation for problem 10:

Step1: Recall the procedure for finding absolute extrema

To find the absolute maximum and minimum of a function $y = h(x)$ on a closed interval $[a,b]$, first find the critical numbers of $h(x)$ in the open interval $(a,b)$, then evaluate $h(x)$ at the critical numbers and at the endpoints $x = a$ and $x = b$.

Step2: Differentiate $h(x)=2x^{3}+3x^{2}-72x + 3$

Using the power rule, $h^{\prime}(x)=6x^{2}+6x - 72$.

Step3: Find the critical numbers

Set $h^{\prime}(x)=0$. Factor out a 6: $6(x^{2}+x - 12)=0$. Then factor the quadratic: $6(x + 4)(x - 3)=0$. Solving $(x + 4)(x - 3)=0$ gives $x=-4$ and $x = 3$. But $x=-4\notin[0,5]$, so we only consider $x = 3$.

Step4: Evaluate $h(x)$ at critical number and endpoints

$h(0)=2(0)^{3}+3(0)^{2}-72(0)+3 = 3$. $h(3)=2(3)^{3}+3(3)^{2}-72(3)+3=2\times27 + 3\times9-216 + 3=54 + 27-216+3=-132$. $h(5)=2(5)^{3}+3(5)^{2}-72(5)+3=2\times125+3\times25 - 360+3=250 + 75-360 + 3=-32$.

Answer for problem 10:

The absolute maximum of $h(x)$ on $[0,5]$ is $h(0)=3$ and the absolute minimum is $h(3)=-132$.