Question

determine whether rolles theorem applies to the given function on the given interval. if so, find the point(s) that are guaranteed to exist by rolles theorem.

$f(x)=3\tan(2x);\left\frac{\pi}{4},\frac{3\pi}{4}\
ight$

since $f$ is not continuous on the closed interval $\left\frac{\pi}{4},\frac{3\pi}{4}\
ight$, not differentiable on the open interval $\left(\frac{\pi}{4},\frac{3\pi}{4}\
ight)$ and $f\left(\frac{\pi}{4}\
ight)\
eq f\left(\frac{3\pi}{4}\
ight)$, rolles theorem cannot be applied to $f$ on the interval $\left\frac{\pi}{4},\frac{3\pi}{4}\
ight$.

Explanation:

Step1: Check continuity

The function \(y = \tan t\) has vertical asymptotes at \(t=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\). For \(f(x)=3\tan(2x)\), when \(x=\frac{\pi}{2}\in[\frac{\pi}{4},\frac{3\pi}{4}]\) (since \(2x=\pi\) when \(x = \frac{\pi}{2}\)), \(\tan(2x)\) is undefined. So \(f(x)\) is not continuous on \([\frac{\pi}{4},\frac{3\pi}{4}]\).

Step2: Check differentiability

Since \(f(x)\) is not continuous on \([\frac{\pi}{4},\frac{3\pi}{4}]\), it cannot be differentiable on \((\frac{\pi}{4},\frac{3\pi}{4})\) (a function must be continuous on a closed interval \([a,b]\) to be differentiable on the open interval \((a,b)\)). Also, \(f(\frac{\pi}{4})=3\tan(\frac{\pi}{2})\) (undefined) and \(f(\frac{3\pi}{4})=3\tan(\frac{3\pi}{2})\) (undefined). But even if we consider the limit - like behavior, the non - continuity and non - differentiability (due to the asymptote in the interval) violate the conditions of Rolle's Theorem.

Answer:

Since \(f\) is not continuous on the closed interval \([\frac{\pi}{4},\frac{3\pi}{4}]\), not differentiable on the open interval \((\frac{\pi}{4},\frac{3\pi}{4})\) and \(f(\frac{\pi}{4})
eq f(\frac{3\pi}{4})\) (in the sense of non - existence of finite values), Rolle's Theorem cannot be applied to \(f\) on the interval \([\frac{\pi}{4},\frac{3\pi}{4}]\).