For what values of x, if any, does #f(x) = 1/(xe^x-3) # have vertical asymptotes?

1 Answer
Jan 28, 2016

Answer:

There is a vertical asymptote at the solution to #xe^x=3#

Explanation:

If there is a solution to #xe^x=3#, call it #c#, then there is a vertical asymptote at #x= c#, because the limit as #x rarr c# of #f(x)# will have the form #1/0#, so the function is either increasing or decreasing without bound as #x= c#.

It is clear that we can make #xe^x < 3# (for example, by making #x=0#).

We can also make #xe^x > 3# (for example, by making #x=10#)

We also know that #xe^x# is continuous on #(-oo,oo)#, so by the Intermediate Value Theorem, there is a solution to #xe^x=3#.

So we know that there is a vertical asymptote, and we can describe where it occurs.

Unfortunately there is not a nice expression for the solution to #xe^x=3#. Using numerical or technological methods we can get an approximation of #c ~~ 1.05#