# How do we use Rolle's Theorem to find whether function f(x)=4x^2-8x+7 has a point where f'(x)=0 in the interval [-1,3]?

Mar 31, 2017

$x = 1$, for details please see below.

#### Explanation:

Rolle's Theorem when applied the function $f \left(x\right)$ must be continuous for $x$ in the given range, here $\left[- 1 , 3\right]$ and $f \left(x\right)$ must be differentiable for $x$ in $\left(- 1 , 3\right)$.

Here we have $a = - 1$ and $b = 3$ and as $f \left(x\right) = 4 {x}^{2} - 8 x + 7$, we have $f \left(- 1\right) = 19$ and $f \left(3\right) = 19$ and hence $f \left(a\right) = f \left(b\right)$.

Now according to Rolle's Theorem, if in such a function $f \left(a\right) = f \left(b\right)$, there is one $c$, where $f ' \left(c\right) = 0$

As $f ' \left(x\right) = 8 x - 8$ and as $8 x - 8 = 0 \implies x = 1$, we have at $x = 1$, $f ' \left(1\right) = 0$

and our $c$ is $1$.

graph{(y-4x^2+8x-7)((x+1)^2+(y-19)^2-0.01)((x-3)^2+(y-19)^2-0.01)((x-1)^2+(y-3)^2-0.015)=0 [-4, 4, -5, 23]}