# For what values of x, if any, does f(x) = 1/((12x-9)sin(pi+(3pi)/x)  have vertical asymptotes?

Oct 1, 2016

$x = 0 , \frac{3}{k} , k \in \mathbb{Z}$ (where $k$ is an integer)

#### Explanation:

Note that for a rational function such as the given function $f$, there will be a rational function whenever its denominator equals $0$.

So, there is a vertical asymptote whenever $\left(12 x - 9\right) \sin \left(\pi + \frac{3 \pi}{x}\right) = 0$.

We can split this into two parts:

$\left\{\begin{matrix}12 x - 9 = 0 \\ \sin \left(\pi + \frac{3 \pi}{x}\right) = 0\end{matrix}\right.$

The first can be solved to show that $x = \frac{9}{12} = \frac{3}{4}$.

The second is a little more difficult: note that $\sin \left(x\right) = 0$ when $x = 0 , \pi , 2 \pi$, and so on. This can be written as $x = k \pi$, where $k \in \mathbb{Z}$, which means $k$ is an integer.

Thus:

$\pi + \frac{3 \pi}{x} = k \pi \text{ "" "," " } k \in \mathbb{Z}$

Subtracting $\pi$, we see that the right hand side remains $k \pi$, because some integer multiple of $\pi$ minus $\pi$ is still an integer multiple of $\pi$.

$\frac{3 \pi}{x} = k \pi \text{ "" "," " " } k \in \mathbb{Z}$

Rearranging:

$3 \pi = x \left(k \pi\right) \text{ "" "," "" } k \in \mathbb{Z}$

$x = \frac{3 \pi}{k \pi} = \frac{3}{k} \text{ "" "," "" } k \in \mathbb{Z}$

Note that the previous solution from $12 x - 9 = 0$ gave $x = \frac{3}{4}$, which is included in the solution $x = \frac{3}{k} , k \in \mathbb{Z}$.

Furthermore, since $\frac{3 \pi}{x}$ is included within the sine function, there will be a vertical asymptote at $x = 0$.