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

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Answer 1 · 2015-11-25T06:43:30

#f# is decreasing at #x=2#.

#f'(x)=frac{d}{dx}(frac{x-e^x}{x-x^2})#

#=frac{(x-x^2)frac{d}{dx}(x-e^x)-(x-e^x)frac{d}{dx}(x-x^2)}{(x-x^2)^2}#

#=frac{(x-x^2)(1-e^x)-(x-e^x)(1-2x)}{x^2(1-x)^2}#

#=frac{(cancel(x)-x^2-xe^x+x^2e^x)-(cancel(x)-e^x-2x^2+2xe^x)}{x^2(1-x)^2}#

#=frac{-3xe^x+x^2e^x+e^x+x^2}{x^2(1-x)^2}#

#f'(2)=1-frac{e^2}{4}<0#

Therefore, #f# is decreasing at #x=2#.