Consider the following function and closed interval.

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

Is $f$ continuous on the closed interval $[-2,2]?$

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

Yes, $f$ is continuous 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 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.

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

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

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

$f(-2)=$

$f(2)=$

Find $\frac{f(b)-f(a)}{b-a}$ 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 $\frac{f(b)-f(a)}{b-a}$ is not defined.

If the mean value theorem can be applied, find all values of $c$ that satisfy the conclusion of the mean value theorem. $($Enter your answers as a comma$-$separated list. If it does not satisfy the hypotheses, enter DNE$).$

$c=$