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

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Answer 1 · 2017-01-18T13:44:40

f(x) is increasing at x=2

First, let's differentiate f(x):

#f'(x)=((1-e^x-xe^x)(x-1)-(x-xe^x))/(x-1)^2#

#=(cancelx-1cancel(-xe^x)-e^x-x^2e^xcancel(+xe^x)cancel-x+xe^x)/(x-1)^2#

#=(-1-e^x-x^2e^x+xe^x)/(x-1)^2#

Then, let's calculate

#f'(2)=(-1-e^2-4e^2+2e^2)/9#

#=(-1-3e^2)/9<0#

Then f(x) is increasing at x=2

{ graph{(x-xe^x)/(x-1) [-20, 20, -227.2, 124.8]}