已知f(x)=(x+1)|x-1|,若关于x的方程f(x)=x+m有三个不同的实数解,求实

已知f(x)=(x+1)|x-1|,若关于x的方程f(x)=x+m有三个不同的实数解,求实数m的取值范围.

答案

思路分析:先分类讨论去绝对值符号,再画出图像辅助求解.

解:

由f(x)=(x+1)|x-1|=画出函数y=f(x)的图像,如图所示,

得x2+x+m-1=0,∵Δ=1-4(m-1)=5-4m,由Δ=0,得m=

由图像得实数m的取值范围是-1<m<,即(-1,).

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