What are the asymptote(s) and hole(s), if any, of $f(x) =1/(e^x-2)$ ?

Question

1822 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-13T15:12:20

There is a asymptote at $x=ln2~~0.6931$

There are no holes because no factors will cancel out of the numerator and denominator, the asymptote(s) are when we have $1/0$

$e^x-2=0$

$e^x=2$

$ln(e^x)=ln2$

$x=ln2$

graph{1/(e^x-2) [-10, 10, -5, 5]}

What are the asymptote(s) and hole(s), if any, of f(x) =1/(e^x-2) f(x) =1/(e^x-2)?