## Calculus 10th Edition

$c=2.$
Since $f(x)$ is a polynomial, it is continuous for all values of $x.$ $f(0)=(0)^2-6(0)+8=8.$ $f(3)=(3)^2-6(3)+8=-1.$ Since $f(x)$ is continuous over the interval, and $-1\lt0\lt8,$ then by the Intermediate Value Theorem, there exists a value $c$ in the interval $[0, 3]$ such that $f(c)=0.$ $x^2-6x+8=0\to(x-2)(x-4)=0\to x=2$ or $x=4$ $c=2$ (since $4$ is rejected as it is not in the interval).