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已知函数f(x0=1\3x^3-a\2x^2+2x+1,且x1,x2是f(x)的两个极点,0<x1<1<x2<3,
若|x1-x2|大于等于m^2-2bm-2,对b属于-1到1恒成立,求m范围
人气:330 ℃ 时间:2019-10-18 08:38:07
解答
f(x)=1\3x^3-a\2x^2+2x+1,f‘(x)=x^2-ax^2+2,x=[a±√(a²-8)]/2,∵0<x1<1<x2<3,∴0<[a-√(a²-8)]/2<1,1<[a+√(a²-8)]/2<3,得3<a<11/3.∵|x1-x2|=√(a²-8)],∴1<|x1-x2|<7/3,∵|x1-x...
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