# For what values of x, if any, does #f(x) = 1/((x-3)(x+2)(e^x-x)) # have vertical asymptotes?

##### 1 Answer

#### Explanation:

The function will have a vertical asymptote wherever the denominator is

In this case, our job is a little easier since we don't have to check whether or not the numerator is

The denominator will be

#x - 3 = 0#

#color(red)(x = 3#

#x + 2 = 0#

#color(red)(x = -2#

#e^x - x = 0#

#e^x = x# Hmm... this might be trickier. Does

#e^x=x# have any solutions?Let's consider if

#x# is negative. Since#e^x# is always positive, it cannot equal#x# if#x# is negative. So we have no negative solutions.If

#x# is 0, then#e^x = 1# , so that is not a solution.This leaves us with the fact that

#x# could be positive, but we need to consider how fast the functions are increasing.For any value of

#x# ,#e^x# is increasing at a rate of#e^x# , and#x# is increasing at a rate of#1# . For positive values of#x# ,#e^x > 1# . This means that for positive values of#x# ,#e^x# increases faster than#x# .Since

#e^x# starts out (at#x=0# ) greater than#x# , and continues to grow faster than it, it will ALWAYS be greater than#x# for any value, since the function#y=x# can never "catch up" to it.Therefore, there are no positive solutions to

#e^x = x# .This means there are no solutions to

#e^x = x# .

Therefore, the function

*Final Answer*