题目
已知一次函数f(x)=(m^2-1)x+m^2-3m+2,若f(x)是减函数,且f(x)=0, 若f(x)≥x^2,求x的取值范围
提问时间:2020-12-15
答案
∵一次函数f(x)=(m^2-1)x+m^2-3m+2,若f(x)是减函数,且f(x)=0
∴m^2-1=x^2
即-3/4x+3/4>=x^2
4x^2+3x-3
∴m^2-1=x^2
即-3/4x+3/4>=x^2
4x^2+3x-3
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