# How do you verify whether rolle's theorem can be applied to the function f(x)=1/x^2 in [-1,1]?

Dec 24, 2016

$f \left(x\right)$ does not satisfy the conditions of Rolle's Theorem on the interval $\left[- 1 , 1\right]$, as $f \left(x\right)$ is not continuous on the interval

#### Explanation:

Rolle's Theorem states that if a function, $f \left(x\right)$ is continuous on the closed interval $\left[a , b\right]$, and is differentiable on the interval, and $f \left(a\right) = f \left(b\right)$, then there exists at least one number $c$, in the interval such that $f ' \left(c\right) = 0$.

So what Rolle's Theorem is stating should be obvious as if the function is differentiable then it must be continuous (as differentiability $\implies$ continuity), and if its is continuous and $f \left(a\right) = f \left(b\right)$ then the curve must chge direction (at least once) so it must have at least one minimum or maximum in the interval.

With $f \left(x\right) = \frac{1}{x} ^ 2$ with $x \in \left[- 1 , 1\right]$, then we can see $f \left(1\right) = 1 = f \left(- 1\right)$, so our quest to verify Rolle's Theorem is to find a number $c$ st $f ' \left(c\right) = 0$.

Differentiating wrt $x$ we have, $f ' \left(x\right) = - 2 {x}^{-} 3 = - \frac{2}{x} ^ 3$

To find a turning point we require;

$f ' \left(x\right) = 0 \implies - \frac{2}{x} ^ 3 = 0$

Which has no finite solution. We can conclude that $f \left(x\right)$ does not satisfy the conditions of Rolle's Theorem on the interval $\left[- 1 , 1\right]$, so it must be that $f \left(x\right)$ is not continuous in that interval.

We can see that this is the case graphically, as f(x) has a discontinuity when $x = 0 \in \left[- 1 , 1\right]$:

graph{1/x^2 [-10, 10, -2, 10]}