## Calculus 10th Edition

$c=3.$
Since $f(x)$ is a polynomial, it is continuous for all values of $x.$ $f(0)=(0)^2+0-1=-1.$ $f(5)=(5)^2+5-1=29.$ Since f(x) is continuous and $-1\lt 11\lt29,$ then there exists a value $c$ in the interval $[0, 5]$ such that $f(c)=11.$ $x^2+x-1=11\to x^2+x-12=0\to$ $(x+4)(x-3)\to x=-4$ (rejected; not in interval) or $x=3.$ $c=3.$