differentiation

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

For 0≦x≦2,
f'(x)=0 when x=ln2.
When x=ln2,
f(x)=(ln2)e^(ln2)-(ln2)^2-2ln2
=2ln2-2ln2-2ln2
=-2ln2
The extrema is (ln2,-2ln2).

Since lim(x->ln2-) f'(x) is -ve,
lim(x->ln2+) f'(x) is +ve,
(ln2,-2ln2) is a minimum point of the function.





2010-11-29 19:00:55 補充：
As (ln2,-2ln2) is the only minmum point when 0≦x≦2,
The global maxima occurs at either x=0 or x-2.
When x=0,f(x)=0
When x=2,f(x)=2e^2-8
As 2e^2-8>0,
The global maxima occurs at x=2 and equals to 2e^2-8.
The global minima occurs at x=ln2 and equals to -2ln2, or -(ln2)^2

Hope the above can help you~

2010-11-29 19:02:11 補充：
(ln2,-2ln2) is the only minmum point when 0<=x<=2

2010-11-29 19:11:18 補充：
f '(x)=(x+1)(e^x-2)
For 0<=x<=2,
f '(x)>0 when x>ln2
f' '(x)<0 when x





2010-11-29 19:20:49 補充：
Better explanation of global maximum:
f '(x)=(x+1)(e^x-2)
For 0<=x<=2,
f '(x)>0 when x>ln2
f '(x)<0 when x




2010-11-29 19:23:56 補充：
Better explanation of global maximum:
f '(x)=(x+1)(e^x-2)
For 0<=x<=2,
f '(x)>0 when x>ln2
f '(x)<0 when x smaller than ln2
f is decreasing at 0<=x<=ln2
f is increasing at ln2<=x<=2
Global maximum occurs at x=0 or x=2