determine whether rolles theorem applies to the given function on the given interval. if so, find the…

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.\n\n$f(x)=3\\tan(2x);\\left\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right$\n\nsince $f$ is not continuous on the closed interval $\\left\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right$, not differentiable on the open interval $\\left(\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right)$ and $f\\left(\\frac{\\pi}{4}\\right)\\neq f\\left(\\frac{3\\pi}{4}\\right)$, rolles theorem cannot be applied to $f$ on the interval $\\left\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right$.

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.\n\n$f(x)=3\\tan(2x);\\left\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right$\n\nsince $f$ is not continuous on the closed interval $\\left\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right$, not differentiable on the open interval $\\left(\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right)$ and $f\\left(\\frac{\\pi}{4}\\right)\\neq f\\left(\\frac{3\\pi}{4}\\right)$, rolles theorem cannot be applied to $f$ on the interval $\\left\\frac{\\pi}{4},\\frac{3\\pi}{4}\\right$.

Answer

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})\neq 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}]).