## Calculus 10th Edition

$c=2.$
Since $f(x)$ is a polynomial, it is continuous for all values of $x.$ $f(0)=0^3-(0)^2+0-2=-2.$ $f(3)=3^3-(3)^2+3-2=19.$ Since $f(x)$ is continuous over the interval $[0, 3]$ and $-2\lt4\lt19,$ then the Intermediate Value Theorem guarantees the existence of a value $c$ such that $f(c)=4.$ $x^3-x^2+x-2=4\to x^3-x^2+x-6=0\to$ $(x-2)(x^2+x+3)=0\to x=2\to c=2.$