Consider the following function and dosed interval.

$f(x)=x^3-3x3,[-2,2]$

Is $f$ continuous on the dosed interval $[-2,2\}?$

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

Yes, $f$ is centinuous on $-2,2$ and differentiable on $(-2,2)$ since polynomials are continuous and differentiable on $R.$

No$,f$ is not continuous on $-2,2.$

No$,f$ is centinusus on $-2,2$ but not differentiable on $(-2,2).$

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

If $r$ is differentiable on the open interval $(-2,2),$ find $r^{\text{'}}(x).($If it is not differentiable on the open interval, enter Dive.$)$

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

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

$(x-2)=$

$(2)=$

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

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

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

Yes, the Mean Value Theorem can be applied.

No$,$ because $f$ is not continuous on the closed interval $\{-2,2\}.$

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

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

$c=$