given the function g(x) = tan x, determine the location of the asymptote on the interval (-3π, -2π).\nox =…

Explanation:

Step1: Recall tangent - asymptote formula

The vertical asymptotes of (y = \tan x) occur at (x=\frac{\pi}{2}+n\pi), where (n\in\mathbb{Z}).

Step2: Find (n) for the given interval

We want to find (n) such that (-3\pi<\frac{\pi}{2}+n\pi < - 2\pi). First, solve the left - hand inequality (-3\pi<\frac{\pi}{2}+n\pi): (-3\pi-\frac{\pi}{2}<n\pi), (-\frac{6\pi + \pi}{2}<n\pi), (-\frac{7\pi}{2}<n\pi), (n>-\frac{7}{2}=-3.5). Then, solve the right - hand inequality (\frac{\pi}{2}+n\pi < - 2\pi): (n\pi<-2\pi-\frac{\pi}{2}), (n\pi<-\frac{4\pi+\pi}{2}), (n\pi<-\frac{5\pi}{2}), (n<-\frac{5}{2}=-2.5). Since (n\in\mathbb{Z}), (n = - 3).

Step3: Calculate the asymptote

Substitute (n=-3) into (x=\frac{\pi}{2}+n\pi). (x=\frac{\pi}{2}+(-3)\pi=\frac{\pi - 6\pi}{2}=-\frac{5\pi}{2}).

Answer:

(x =-\frac{5\pi}{2})