# For what values of x, if any, does f(x) = 1/((5x+8)cos(pi/2-(12pi)/x)  have vertical asymptotes?

Jul 5, 2017

$x = - \frac{8}{5}$, $x = 0$, and $x = - \frac{6}{n}$ (see explanation)

#### Explanation:

For there to be a vertical asymptote, the denominator must equal to zero, and $0 \cdot \text{anything} = 0$, so $\left(5 x + 8\right) = 0 , 5 x = - 8 , x = - \frac{8}{5}$.

Also, $\cos \left(\left(\frac{\pi}{2}\right) - \frac{12 \pi}{x}\right) = 0$. ${\cos}^{- 1} \left(0\right) = \frac{\pi}{2} + n 2 \pi$.

Therefore, $\frac{\pi}{2} - \frac{12 \pi}{x} = \frac{\pi}{2} + n 2 \pi$, $\frac{12 \pi}{x} = - n 2 \pi$, $x = - \frac{12 \pi}{n 2 \pi}$, $x = - \frac{6}{n}$. By rearranging we get $n = - \frac{6}{x}$. We can substitute this to get $\frac{\pi}{2} - \frac{12 \pi}{x} = \frac{\pi}{2} + 2 \left(- \frac{6}{x}\right) \pi \equiv \frac{\pi}{2} - \frac{12 \pi}{x} = \frac{\pi}{2} - \frac{12}{x} \pi$.

For $n$ to be an integer, $2 n \equiv - \frac{12}{x}$, where $x$ is any number which 12 is divisible by, to also give a whole number. For example, let's take $x = 4 , 12 \text{ and } 24$:

4:
$2 n = - \frac{12}{4} = - 3$, as $2 n$ is an integer, $\frac{\pi}{2} - \frac{12 \pi}{x}$ can equal $\frac{\pi}{2} + 2 \left(- \frac{6}{x}\right) \pi$, $\frac{\pi}{2} - 3 \pi = \frac{\pi}{2} - 3 \pi$, $\cos \left(\frac{\pi}{2} - 3 \pi\right) = 0$, $0 \left(5 \left(- 3\right) + 8\right) = 0$.

12:
$2 n = - \frac{12}{12} = - 1$, as $2 n$ is an integer, $\frac{\pi}{2} - \frac{12 \pi}{x}$ can equal $\frac{\pi}{2} + 2 \left(- \frac{6}{x}\right) \pi$, $\frac{\pi}{2} - 1 \pi = \frac{\pi}{2} - 1 \pi$, $\cos \left(\frac{\pi}{2} - \pi\right) = 0$, $0 \left(5 \left(- 1\right) + 8\right) = 0$.

24:
$2 n = - \frac{12}{24} = - \frac{1}{2}$, as $2 n$ is not an integer, $\frac{\pi}{2} - \frac{12 \pi}{x}$ cannot equal $\frac{\pi}{2} + 2 \left(- \frac{6}{x}\right) \pi$, as only $\cos \left(\frac{\pi}{2} + 2 n \pi\right)$ can equal 0, where $2 n$ forms an integer, but $- \frac{1}{2}$ is not an integer. For example, $\cos \left(\frac{\pi}{2}\right) = 0$, $\cos \left(\frac{\pi}{2} + 2 \pi\right) = 0$, and $\cos \left(\frac{\pi}{2} + 4 \pi\right) = 0$.

Also, $x = 0$ is an asymptote as $\frac{12 \pi}{0} = \text{undefined}$.