
Forget the calculus, sketch a graph.
posted by Damon

10 2.0005
3 2.14
2 2.27
1 2.4 maximum
0.5 2.3
0 2
1 .72
2 12.8
10 220,263 minimum as x >oo
derivative
y' = [ xe^x + e^x]
= e^x [x+1]
so max or min at x = 1
second derivative
y" =  [e^x(1)+(x+1)e^x]
= e^x [ x+2 ]
if x = 1
then = (1/e)(1)
negative so maximumposted by Damon
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