Is #f(x)=(x-e^x)/(x-1)^3# increasing or decreasing at #x=2#?

1 Answer
Sep 29, 2017

Increasing.

Explanation:

To know if a function #f(x)# is increasing or decreasing at some point, you must find the derivative #f'(x)# of the function and check the sign of the value of the derivative. If #f'(x) > 0#, the function is said to be increasing. If #f'(x) < 0#, the function is said to be decreasing.

Begin by using the Quotient Rule to find the derivative #f'(x)#:

#f'(x) = ( (x-1)^3 * (1 - e^x) - (x - e^x) * (3 * (x-1)^2) ) / ((x-1)^3)^2 #
# = ( (x-1)^3 * (1 - e^x) - 3(x - e^x)(x-1)^2 ) / (x-1)^6 #

Now we can evaluate #f'(2)#:

#f'(2) = ( (2-1)^3 * (1 - e^2) - 3(2 - e^2)(2-1)^2 ) / (2-1)^6 #
# = ( (1 - e^2) - 3(2 - e^2) ) / 1 = (1 - e^2 - 6 +3e^2 ) #
# = 2e^2 - 5 ~~ 9.778 > 0#

Since #f'(2) > 0#, we say #f(x)# is increasing at #x = 2#.

This can also be verified by looking at the graph as a support for our work:

graph{(x-e^x)/(x-1)^3 [-5.484, 12.296, -7.46, 1.425]}