For what values of x, if any, does f(x) = 1/((3x-2)sin(pi+(2pi)/x)  have vertical asymptotes?

Mar 23, 2017

$x = \left\{- 2 , \frac{2}{3}\right\}$

Explanation:

Vertical symptotes occur where f(x) is undefined. This will occur where the function is over 0 in this case. We must therefore determine when the denominator equals 0. Whenever two things are multiplied together, such as in this case $3 x - 2$ and $\sin \left(\pi + \frac{2 \cdot \pi}{x}\right)$, the total will be equal to zero when either or both are zero so we must find when both equal 0. To do this, set each expression separately equal to 0 and solve for x.

Now we have the equations $3 x - 2 = 0$ and $\sin \left(\pi + \frac{2 \cdot \pi}{x}\right) = 0$. The former is much easier to solve. Simply add two to both sides and divide both sides by three to get: $x = \frac{2}{3}$.

For the next equation we must first find where sine is equal to 0. This only occurs at the quadrantals of 0, $\pi$, and $2 \cdot \pi$ within a range of 0 to $2 \cdot \pi$ for the angle. Since $2 \cdot \pi$ and 0 are the same angle we can disregard one. This means that sine is equal to 0 whenever its argument is 0 or $\pi$. From this we can make two MORE equations from setting the argument equal to these two things.

This gives us: $\pi + \frac{2 \cdot \pi}{x} = 0$ and $\pi + \frac{2 \cdot \pi}{x} = \pi$. For the second equation subtract $\pi$ from both sides and multiply both sides by x to get: $2 \cdot \pi = 0$ which means that this case can never happen because $2 \pi$ will NEVER equal 0 and we can disregard it. For the first equation factor out $2 \pi$ from the left expression to get $2 \pi \cdot \left(\frac{1}{2} + \frac{1}{x}\right) = 0$ then divide by $2 \pi$ on both sides to get $\frac{1}{2} + \frac{1}{x} = 0$. Subtract $\frac{1}{2}$ from both sides and multiply by x on both sides to get $1 = - \frac{x}{2}$ then multiply by $- 2$ on both sides to finally get: $x = - 2$.

Finally if we combine all of the x's we found we get that there are vertical asymptotes for this function at $x = \left\{- 2 , \frac{2}{3}\right\}$ within a range of 0 to $2 \pi$. Hope I helped!