Answer
Rolle's Theorem can be applied; $c=\dfrac{6+\sqrt{3}}{3}$ or $c=\dfrac{6-\sqrt{3}}{3}.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x$.
$f(1)=f(3)=0.$
Since $f(x)$ is continuous over $[1 , 3]$ and differentiable over $(1, 3)$, applying Rolle's Theorem over the interval $[1, 3]$ guarantees the existence of at least one value $c$ such that $1\lt c\lt3 $ and $f'(c)=0.$
Applying the product rule for three terms gives us the derivative to be:
$f'(x)=(1)(x-2)(x-3)+(x-1)(1)(x-3)+(x-1)(x-2)(1)$
$=3x^2-12x+11\to f'(x)=0\to 3x^2-12x+11=0\to$
Using the Quadratic Equation gives us $c=\dfrac{6+\sqrt{3}}{3}$ or $c=\dfrac{6-\sqrt{3}}{3}.$
