What are the points of inflection, if any, of #f(x)=e^2x - e^x #?

1 Answer
Sep 28, 2017

There are no points of inflection.

Explanation:

To find points of inflection, you must first find critical points of the second derivative #(d^2y)/(dx^2)# (hereafter referred to as #y''# for convenience. These points are #x# values for which #y''(x)# has a value of 0, or is undefined. You then must examine the concavity of the function "around" those points to determine if the concavity changes "at" those points.

Begin by finding the first derivative #dy/dx# (referred to as #y'#) using basic derivative rules. (Note: Despite it's appearance, #e^2# is a constant value and #bb("not")# a function.)

#y' = e^2 - e^x#

(This is because #e^2x# behaves like #Cx# for differentiation.)

Next, the 2nd derivative #y''#:

#y'' = -e^x#

(The #e^2# term is a constant.)

The function #-e^x# is a "well-behaved" function that has no points where it is undefined. Thus, the only critical points for #y''# would come when #-e^x = 0#. However, there is no real number solution for this equation.

Since there are no critical points for the second derivative, and since #y'' < 0# for all #x#, there are no points of inflection.

graph{(e^2)x - e^x [-32.47, 32.48, -16.22, 16.26]}