Question

How do you verify whether Rolle's theorem can be applied to the function $f(x) = \tan x$ in $\left[ {0,pi} \right]$?

Answer 1

Hint: Rolle’s theorem states that if a function $f(x)$ is continuous in the interval $[a,b]$, differentiable in the interval $(a,b)$ and $f(a) = f(b)$, then $f'(x) = 0$ for some value of $x$ in the interval $[a,b]$. Rolle’s theorem can be applicable on a function if it satisfies all the condition mentioned, i.e. $f(x)$ is continuous in the interval $[a,b]$, differentiable in the interval $(a,b)$ and $f(a) = f(b)$.

Complete step by step solution:
We have been given to check whether rolle’s theorem is applicable on the function $f(x) = \tan x$ in the interval $\left[ {0,pi} \right]$.
We will check each condition for rolle’s theorem one-by-one to check for its applicability.
First condition states that the function $f(x)$ should be continuous in the interval $[a,b]$, i.e. $\tan x$ should be continuous in the interval $\left[ {0,pi} \right]$.
We know that the value of $\tan x$ is not defined at $x = \dfrac{\pi }{2}$. So the function $\tan x$ is not continuous in the interval $\left[ {0,pi} \right]$. Thus, the first condition does not hold true.
We can say at this point itself that Rolle's theorem is not applicable to the given function in the given interval. But, we will also check other conditions for clarification.
Second condition mentions that the function $f(x)$ should be differentiable in the interval $(a,b)$, i.e. $\tan x$ should be differentiable in the interval $\left( {0,pi} \right)$.
Since, $\tan x$ is not defined at $x = \dfrac{\pi }{2}$, $\tan x$ is not continuous in the interval $\left( {0,pi} \right)$. So it is not differentiable in the interval $\left( {0,pi} \right)$.
Thus, the second condition also does not hold true.
Third condition mentions that $f(a) = f(b)$, i.e. $\tan 0$ should be equal to $\tan \pi $. We know that $\tan 0 = \tan \pi = 0$. Thus, only the third condition holds true.
Since, two out of three conditions does not hold true, we can say that rolle’s theorem is not applicable to the function $f(x) = \tan x$ in the interval $\left[ {0,pi} \right]$.

Note: While checking for the applicability for the rolle’s theorem on a function we can check the applicability for a closed interval only. For the conditions, we check for continuity of the function in the closed interval and differentiability in the open interval. If even any one of the conditions is not satisfied, we cannot apply Rolle's theorem on the function in the given interval.