二次函数f(x)=ax^2+bx+c的系数a,b,c互不相等
若1/a,1/b,1/c成等差数列,a,b,c成等比数列,且f(x)在[-1,0]的最大值为-6
则a=?
人气:192 ℃ 时间:2019-10-10 01:35:21
解答
如果回答得好,
由题知abc都不等于0
由等差数列知:1/a+1/c=2/b
化简:b(a+c)=2ac
由等比数列知:b^2=ac带入上式
得:a+c=2b
又由b^2=ac得c=b^2/a带入上式
得:(a-b)^2=0
故a=b
同理a=c
故:a=b=c
f(x)=ax^2+ax+a=a(x^2+x+1)
=a[(x+1/2)^2+3/4]
如果a>0
f(x)的最大值=f(-1)=f(0)=a=-6
又-6
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