Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=e^x/(1-e^(3x^2-x))`?

Answer 1

No discontinuities.

Vertical asymptotes at `x=0` and `x=1/3`

Horizontal asymptote at `y=0`

Explanation:

To find the vertical asymptotes, we equate the denominator to `0`.

Here,

`1-e^(3x^2-x)=0`

`-e^(3x^2-x)=-1`

`e^(3x^2-x)=1`

`ln(e^(3x^2-x))=ln(1)`

`3x^2-x=0`

`x(3x-1)=0`

`x=0, 3x-1=0`

`x=0,x=1/3`

`x=1/3,0`

So we find vertical asymptote is at `x=1/3,0`

To find the horizontal asymptote, we must know one crucial fact: all exponential functions have horizontal asymptotes at `y=0`

Obviously, the graphs of `k^x+n` and other such graphs do not count.

Graphing:

graph{(e^x)/(1-e^(3x^2-x)) [-18.02, 18.03, -9.01, 9.01]}