Answer
Rolle's theorem can be applied; $c=\dfrac{3}{2}.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x$.
$f(0)=f(3)=0.$
Since $f(x)$ is continuous over $[0 ,3 ]$ and differentiable over $(0, 3)$, applying Rolle's Theorem over the interval $[0, 3]$ guarantees the existence of at least one value $c$ such that $0\lt c\lt 3$ and $f'(c)=0.$
$f'(x)=-2x+3\to f'(x)=0 \to c=\dfrac{3}{2}.$
