Question

In Exercises \(7-20,\) find the vertical asymptotes (if any) of the function. $$ f(x)=\tan 2 x $$

Short answer

The function \(f(x)=\tan 2x\) has vertical asymptotes at \(x= \frac{\pi}{4} + \frac{n\pi}{2}\), for all integers \(n\).

Step-by-step solution

Understanding the Function

Firstly, understand that a vertical asymptote of a function is a vertical line \(x=a\) where the function \( f(x) \) tends to infinity as \( x \) tends to \( a \). To find the vertical asymptote of the tangent function, one must know that the tangent of any angle is defined as \(\tan \theta =\frac{\sin \theta}{\cos \theta }\). Since you can't divide by zero, there is a vertical asymptote wherever the cosine function \( \cos \theta \) equals zero.

Identify Where Cosine Equals Zero

The cosine function equals zero at \(\frac{\pi}{2} + n\pi\), where \( n \) is an integer. Since the argument of the tangent function is \(2x\), you need to solve the equation \(2x=\frac{\pi}{2} + n\pi\).

Solve for x

Solving the equation \(2x=\frac{\pi}{2} + n\pi\) for \(x\) gives \(x=\frac{\pi}{4} + \frac{n\pi}{2}\), which is the locus of the vertical asymptotes.