## Calculus 10th Edition

Rolle's Theorem can be applied; $c=\dfrac{6+\sqrt{3}}{3}$ or $c=\dfrac{6-\sqrt{3}}{3}.$
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 $c=\dfrac{6-\sqrt{3}}{3}.$