Question

Vertical asymptote of f(x) = tan2x ? How would you solve to determine it?

Answer 1

Hint: Use the definition of \(\tan(x)\), and see where \(\cos(2x)=0\).

Answer 2

sin/cos?

Answer 3

Eeeyup.

Answer 4

@LolWolf I'm getting like .78 . Do i just solve by Taking the inverse cos of 0 and dividing by 2?

Answer 5

I'm a little rusty on this

Answer 6

Nah, try to find it analytically, use the unit circle to get a general result.

Answer 7

Think you can help me out with that? lol

Answer 8

@LolWolf

Answer 9

Sure. All right, using this:
http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/300px-Unit_circle_angles_color.svg.png
We find that
\[
\cos(x)=0
\]When:
\[
x=\frac{\pi}{2}
\]We also know that, every interval of \(\pi\), the value will repeat (since it is zero for every half-circle), thus, our final solution is:
\[
\cos(x)=0, x=\pi k+\frac{\pi}{2}
\]Where, \(k\in \mathbb{Z}\). Using this, we have that, vertical asymptotes are present in \(\tan(x)\) when \(x\) is the above value.

Answer 10

wait so it isn't sin(2x)/cos(2x)??

Answer 11

I do get the above answer though. My precal knowledge is coming back :) But what happened to the 2x?

Answer 12

Oh, crap, thanks, yes, I forgot about that, haha. So, for that, then:
\[
\cos(2x)=0\implies\\
\cos(y)=0, y=2x\implies x=\frac{y}{2}
\]With this, we simply take the expression above and divide it by two, to find:
\[
x=\frac{2\pi k+\pi}{4}, k\in \mathbb{Z}
\]

Answer 13

So, there we go, that's for \(\tan(2x)\).