已知:f(x)=(x a)ex(x为平方) ½x²,且f′(0)=0,求(1)a=?(2)f(x)的单调区间

f(x)=(x a)e^x ½x²
f'(x)=e^x (x a)e^x x
=(x a 1)e^x x
f'(0)=(0 a 1)*e^0 0
=a 1=0
∴a=-1
(2)
f(x)=(x-1)e^x ½x²
f'(x)=e^x (x-1)e^X x
=xe^x x
=(x 1)e^x
e^x恒>0
∴令f'(x)>=0
x 1>=0
x>=-1
∴f(x)增区间是[-1, ∞)
减区间是(-∞,-1)

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