Question

How do you verify whether rolle’s theorem can be applied to the function $f\left( x \right) = \dfrac{1}{{{x^2}}}$ in $[1, - 1]$ ?

Answer 1

To solve this question, we need to understand the Rolle’s Theorem first. The Rolle's Theorem can be applied to the function $f\left( x \right)$ is continuous on the closed interval $[a,b]$, and is differentiable on the interval, and $f(a) = f(b)$, then there must be at least one number $c$, in the interval such that $f'(c) = 0$. Therefore, we have to check these conditions for the given function in the given interval.

Complete step by step answer:
We will first check the continuity of the function $f\left( x \right) = \dfrac{1}{{{x^2}}}$ in the interval $[1, - 1]$.For that we have to find the value of $f\left( 1 \right)$ and $f\left( { - 1} \right)$.
$f\left( x \right) = \dfrac{1}{{{x^2}}}$
Therefore, $f\left( 1 \right) = \dfrac{1}{{{1^2}}} = 1$ and $f\left( { - 1} \right) = \dfrac{1}{{{{\left( { - 1} \right)}^2}}} = 1$
Now, we will check the differentiability of this function in the given interval. For that, we need to find the value of $x$ such that the value of $f'\left( x \right)$ is zero.
$f'\left( x \right) = 0 \\\ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{{x^2}}}} \right) = 0 \\\ \Rightarrow - \dfrac{2}{{{x^3}}} = 0 \\\$
This has no finite solution. Therefore, We can conclude that $f\left( x \right)$ does not satisfy the conditions of Rolle's Theorem in the interval $[1, - 1]$, so it must be that $f\left( x \right)$ is not continuous in that interval.

Thus, as the given function does not satisfy the conditions of the Rolle’s Theorem in the given function, we cannot apply the theorem to this function for the given interval.

Note: We have seen the conditions of the Rolle’s Theorem. Practically, we can say that if the function is differentiable then it must be continuous. And if it is continuous $f(a) = f(b)$then the curve of the function must change its direction at least once so it must have at least one minimum or maximum value in the interval.