# For what values of x, if any, does f(x) = 1/((9x-5)sin(pi+(6pi)/x)  have vertical asymptotes?

Dec 21, 2017

$x = \frac{5}{9}$

$x = \frac{6}{2 n - 1} \mathmr{and} \frac{3}{n} , n \in \mathbb{Z}$

#### Explanation:

To compute the verticle asymptotes we must consider the values for what the function is undefined at, hence where denominator $= 0$

$\implies \left(9 x - 5\right) \sin \left(\pi + \frac{6 \pi}{x}\right) = 0$

$\implies 9 x - 5 = 0$

$\implies \textcolor{red}{x = \frac{5}{9}}$ This is our first asymptote

To find the second....

$\implies \sin \left(\pi + \frac{6 \pi}{x}\right) = 0$

$\sin \left(\pi + \frac{6 \pi}{x}\right) = \sin \left(0\right)$

$- - - - - - - - - - - -$

Using our knowledge of general solutions...

If $\sin \theta = \sin \gamma$

$\theta = \left\{2 \pi n + \gamma , 2 \pi n + \pi - \gamma\right\}$

$- - - - - - - - - - - -$

$\implies \pi + \frac{6 \pi}{x} = \left\{2 \pi n , 2 \pi n + \pi\right\}$

$\implies \frac{6 \pi}{x} = \left\{2 \pi n - \pi , 2 \pi n\right\}$

$\implies \frac{1}{x} = \left\{\frac{2 n - 1}{6} , \frac{n}{3}\right\}$

$\implies \textcolor{red}{x = \left\{\frac{6}{2 n - 1} , \frac{3}{n}\right\}}$

$n \in \mathbb{Z}$

The graph...

Interesting...