What are the asymptotes and removable discontinuities, if any, of f(x)=2xex2ex?

1 Answer
Jun 9, 2018

asymptote at:

x=ln20.693

No removable discontinuities.

Explanation:

f(x)=2xex2ex

There are no factors that cancel in the numerator and denominator so there are no removable discontinuities (holes). To find the asymptote(s) find when the function is undefined, i.e. 2xex0

2ex=0

ex=2

ln(ex)=ln2

x=ln20.693