Question

What are the asymptotes and removable discontinuities, if any, of `f(x)= 2/( e^(-6x) -4)`?

Answer 1

No removable discontinuities.
Asymptote: `x=-0.231`

Explanation:

Removable discontinuities are when `f(x) = 0/0`, so this function will not have any since its denominator is always 2.

That leaves us finding the asymptotes (where the denominator = 0).

We can set the denominator equal to 0 and solve for `x`.

`e^(-6x)-4=0`
`e^(-6x)=4`
`-6x = ln4`
`x = -ln4/6 = -0.231`

So the asymptote is at `x=-0.231`. We can confirm this by looking at the graph of this function:
graph{2/(e^(-6x)-4) [-2.93, 2.693, -1.496, 1.316]}