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Mathematics 14 Online
OpenStudy (anonymous):

Find the vertical asymptote of f(x)=tan2x

Parth (parthkohli):

\[\tan^2(x)\]or\[\tan(2x)\]?

OpenStudy (anonymous):

tan(2x)

Parth (parthkohli):

Right. The vertical asymptote is just a value of \(x\) which makes \(f(x)\) undefined. Do you follow?

OpenStudy (anonymous):

Yes

Parth (parthkohli):

We also know that \(\tan(\alpha)\) is undefined at \(\alpha = \large{\pi \over 2}\)

Parth (parthkohli):

So, if the vertical asymptote is at \(\tan({\pi \over 2})\), and also at \(\tan(2x)\), then what is x = ?

OpenStudy (anonymous):

Do we set that equal to 0 to solve for x?

OpenStudy (cwrw238):

tan (pi/2) = tan (2x) what do we get if we cross out the tan's?

OpenStudy (anonymous):

2x= pi/2 ?

OpenStudy (cwrw238):

exactly - so solve that for x and u have your answer

OpenStudy (anonymous):

thank you!

OpenStudy (cwrw238):

yw

Parth (parthkohli):

Yeah

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