Answer
See proof
$c = 1$
Work Step by Step
1. $f(x)=2x^{2}-4x+5$
Since this a polynomial, $f(x)$ is continuous at $[-1,3]$ and $f(x)$ is differentiable at $(-1,3)$
2. Now since the first condition is satisfied, prove that $f(a)=f(b)$
$f(-1) = 2\times(-1)^{2}-4\times(-1)+5$
$= 2+4+5$
$= 11$
$f(3) = 2\times(3)^{2}-4\times(3)+5$
$= 18-12+5$
$=11$
Therefore, $f(a)=f(b)$
Moving on to the final step:
$f'(c)=0$
$\frac{d}{dx} (2x^{2}-4x+5) = 0$
$4x-4=0$
$4(x-1)=0$
$c=1$
Thus, we were able to find the value of $c$ that satisfies the conclusions of Rolle's theorem, which is $c =1$.
