Consider the following function and closed interval.

$f(x)=x^3-3x+8,[-2,2]$

$Isf$ continuous $on$ the closed interval $-2,2?$

Yes, $it$ does not matter $iffis$ continuous $or$ differentiable; every function satisfies the mean value theorem.

Yes, $fis$ continuous $on-2,2$ and differentiable $on(-2,2)$ since polynomial are continuous and differentiable $onR.$

$No,fis$ not continuous $on-2,2.$

$No,fis$ continuous $on-2,2$ but not differentiable $on(-2,2).$

There $is$ not enough information $to$ verify $if$ this function satisfies the mean value theorem.

$Iffis$ differentiable $on$ the open interval $(-2,2),$ find $f^{\text{'}}(x).(Ifitis$ not differentiable $on$ the open interval, enter DNE.$)$

$f^{\text{'}}(x)=$

Find $f(-2)$ and $f(2).(Ifan$ answer does not exist, enter DNE.$)$

$f(-2)=$

$f(2)=$

Find $\frac{f(b)-f(a)}{b-a}$ for $[a,b]=[-2,2].(Ifan$ answer does not exist, enter DNE.$)$

$\frac{f(b)-f(a)}{b-a}=$

Determine whether the mean value theorem can $be$ applied $tofon$ the closed interval $-2,2.(Select$ all that apply.$)$

Yes, the Mean Value Theorem can $be$ applied.

$No,$ because $fis$ not continuous $on$ the closed interval $-2,2.$

$No,$ because $fis$ not differentiable $on$ the open interval $(-2,2).$

$No,$ because $\frac{f(b)-f(a)}{b-a}is$ not defined.

$c=$