Answer
The theorem applies
$c=\displaystyle \frac{1}{3}$
Work Step by Step
Rolle's Theorem
Let $f$ be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ .
If $f(a)=f(b)$ , then there is at least one number $c$ in $(a, b)$ such that $f^{\prime}(c)=0.$
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f is continuous on $[-3,2]$ and it is differentiable on $(-3,2)$
Is $f(a)=f(b)?$
$f(-3)=0$
$f(2)=0$
Yes.
The premise of the theorem is satisfied, so we can use its conclusion.
(The theorem applies)
There is a $ c\in (-3,2)$such that $f^{\prime}(c)=0.$
$ f^{\prime}(x)=(1)(x+3)^{2}+(x-2)\cdot[2(x+3)^{1}\cdot 1]\qquad$ ... product rule, chain rule
$=x^{2}+6x+9+2(x^{2}+x-6)$
$=3x^{2}+8x-3$
$=3x^{2}+9x-x-3$
$=3x(x+3)-(x+3)$
$=(x+3)(3x-1)$
$f^{\prime}(x)=0$ for $x=-3\not\in(-3,2)$, and for $x=\displaystyle \frac{1}{3}\in(-3,2).$
Thus, $c=\displaystyle \frac{1}{3}$
