已知函数f(x)=ax^2+bx+1(a不等于0,a,b为实数),设F(x)=｛①f(x)(x＞0)②-f(x)(x＜0)｝.①：_数学解答

已知函数f(x)=ax^2+bx+1(a不等于0,a,b为实数),设F(x)=｛①f(x)(x＞0)②-f(x)(x＜0)｝.①：若f（-1...
已知函数f(x)=ax^2+bx+1(a不等于0,a,b为实数),设F(x)=｛①f(x)(x＞0)②-f(x)(x＜0)｝.
①：若f（-1）=0且对任意实数x均有f(x)＞=0成立,求F(x)表达式.
②：在①的条件下,当x∈[-2,2]时,g(x)=f(x)-kx是单调函数,求实数k的取值范围.
③：设mn＜0,m+n＞0,a＞0,且f(x)满足f(-x)=f(x),试比较F(m)+F(n)与0的大小.

人气：241 ℃　时间：2019-08-19 17:07:09

解答

a-b+1=0b=a+1f(x)=ax^2+(a+1)x+1=(x+1)(ax+1)只有当a=1时,f(x)>=0f(x)=x^2+2x+1F(x)=｛①x^2+2x+1(x＞0)②-x^2-2x-1(x＜0)｝g(x)=x^2+(2-k)x+1g'(x)=2x+2-kg'(-2)g(2)=(-2-k)(6-k)>0k>6或k