Answer:
The asymptotes of [tex]\tan (2x)[/tex] is when [tex]x=\frac{\pi}{4},\frac{3\pi}{4}[/tex]
Step-by-step explanation:
Given : [tex]f(x)=\tan (2x)[/tex] from [tex]x = 0 \text{ to } x = \pi[/tex]
To find : The asymptotes of the function?
Solution :
[tex]f(x)=\tan (2x)[/tex]
Re-written as
[tex]f(x)=\tan (2x)=\frac{\sin (2x)}{\cos (2x)}[/tex]
Now, we need to find the value of x that makes [tex]\cos(2x)=0[/tex]
[tex]\cos(2x)=0[/tex]
[tex]2x=\cos^{-1}0[/tex]
[tex]2x=\frac{\pi}{2},\frac{3\pi}{2}[/tex]
[tex]x=\frac{\pi}{4},\frac{3\pi}{4}[/tex]
The function [tex]\cos(2x)[/tex] has a period of [tex]\pi[/tex]
So, The asymptote of [tex]\tan (2x)[/tex] is when [tex]x=\frac{\pi}{4}+\pi n\text{ or }\frac{3\pi}{4}+\pi n[/tex]
But from [tex]x = 0 \text{ to } x = \pi[/tex]
The asymptotes of [tex]\tan (2x)[/tex] is when [tex]x=\frac{\pi}{4},\frac{3\pi}{4}[/tex]
