# How do you verify whether rolle's theorem can be applied to the function f(x)=tanx in [0,pi]?

Aug 26, 2016

The stickler for precision in me always wants to say, "Yes. For any function at all, Rolle's Theorem say: If $f$ is continuous on $\left[a , b\right]$ and differentiable on $\left(a , b\right)$ and $f \left(a\right) = f \left(b\right)$, then there is a $c$ in $\left(a , b\right)$ with $f ' \left(c\right) = 0$"

When we are asked whether some theorem "can be applied" to some situation, we are really being asked "Are the hypotheses of the theorem true for this situation?"

(The hypotheses are also called the antecedent, of 'the if parts'.)

So we need to determine whether the hypotheses ot Rolle's Theorem are true for the function

$f \left(x\right) = \tan x$ on the interval $\left[0 , \pi\right]$

Rolle's Theorem has three hypotheses:

H1 : $f$ is continuous on the closed interval $\left[a , b\right]$
H2 : $f$ is differentiable on the open interval $\left(a , b\right)$.
H3 : $f \left(a\right) = f \left(b\right)$

We say that we can apply Rolle's Theorem if all 3 hypotheses are true.

H1 : The function $f$ in this problem is NOT continuous on $\left[0 , \pi\right]$ Because it is not defined at $\frac{\pi}{2}$.

At this point we could stop. Not all of the hypotheses are true. Let's look at the others for completeness.

H2 : The function $f$ in this problem is not differentiable on $\left(0 , \pi\right)$ because it is not continuous at $\frac{\pi}{2}$. (The derivative function ${\sec}^{2} x$ is not defined for $x = \frac{\pi}{2}$

H3 : $f \left(0\right) = 0 = f \left(\pi\right)$

If even one hypothesis fails to be true, then we cannot apply the theorem.
Therefore we CANNOT apply Rolle's Theorem to $f \left(x\right) = \tan x$ on the interval $\left[0 , \pi\right]$.