Question

Where are the asymptotes of f(x) = tan 2x from x = 0 to x = π?

Answer 1

tan(x) is undefined when x = pi/2 and x = 3pi/2 for 0 < x < 2pi Therefore, tan(2x) is undefined when x = pi/4 and x = 3pi/4 for 0 < x < pi

Answer 2

The asymptotes of
`\tan (2x)` is when
`x=(\pi)/(4),(3\pi)/(4)`

Explanation:

Given :
`f(x)=\tan (2x)` from
`x = 0 \text{ to } x = \pi`

To find : The asymptotes of the function?

Solution :

`f(x)=\tan (2x)`

Re-written as

`f(x)=\tan (2x)=(\sin (2x))/(\cos (2x))`

Now, we need to find the value of x that makes
`\cos(2x)=0`

`\cos(2x)=0`

`2x=\cos^(-1)0`

`2x=(\pi)/(2),(3\pi)/(2)`

`x=(\pi)/(4),(3\pi)/(4)`

The function
`\cos(2x)` has a period of
`\pi`

So, The asymptote of
`\tan (2x)` is when
`x=(\pi)/(4)+\pi n\text{ or }(3\pi)/(4)+\pi n`

But from
`x = 0 \text{ to } x = \pi`

The asymptotes of
`\tan (2x)` is when
`x=(\pi)/(4),(3\pi)/(4)`