Calculus: Early Transcendentals 8th Edition

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: $f'(c)=4c-4$ $f(c)=0$ therefore, $4c-4=0$ $c=1$ Thus, we were able to find the value of $c$ that satisfies the conclusions of Rolle's theorem.