f'(x) = (1-x)e^-x
f is increasing when f' > 0
e^-x is always positive, so
f is increasing when 1-x > 0
See the graph at
http://www.wolframalpha.com/input/?i=x+e^-x%2C+0%3C%3Dx%3C%3D2
Find the interval on which f(x)=xe^-x is increasing.
I got (1,infinity)..
1 answer