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答:
f(x)=(e^x)(x²+5x-2)
f'(x)=(e^x)(x²+5x-2)+(e^x)(2x+5)
f'(x)=(x²+7x+3)e^x
解f'(x)=0得:
x=(-7±√35)/2
x<(-7-√35)/2或者x>(-7+√35)/2时f'(x)>0
(-7-√35)/2<x<(-7+√35)/2时f'(x)<0
单调递增区间:
( -∞,(-7-√35)/2 )或者( (-7+√35)/2,+∞)
单调递减区间:
( (-7-√35)/2 , (-7+√35)/2)
f(x)=(e^x)(x²+5x-2)
f'(x)=(e^x)(x²+5x-2)+(e^x)(2x+5)
f'(x)=(x²+7x+3)e^x
解f'(x)=0得:
x=(-7±√35)/2
x<(-7-√35)/2或者x>(-7+√35)/2时f'(x)>0
(-7-√35)/2<x<(-7+√35)/2时f'(x)<0
单调递增区间:
( -∞,(-7-√35)/2 )或者( (-7+√35)/2,+∞)
单调递减区间:
( (-7-√35)/2 , (-7+√35)/2)
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