#### Answer

$c=3.$

#### Work Step by Step

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 $x=3.$
$c=3.$