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

1 Answer
Oct 8, 2017

Answer:

#x = 3" "" "# or #" "" "x = -2#

Explanation:

The function will have a vertical asymptote wherever the denominator is #0# and the numerator is not also #0#.

In this case, our job is a little easier since we don't have to check whether or not the numerator is #0#, since it is always #1#.

The denominator will be #0# whenever any of its factors #(x-3)#, #(x+2)#, and #(e^x-x)# are #0#. So, let's set each of them equal to #0# and solve for #x#.

#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 #f(x) = 1/((x-3)(x+2)(e^x-x))# will have asymptotes at #x=3# and #x=-2#.

Final Answer