Question

设m属于R,f(x)=2x^2+(x-m)|x-m|,g(x)={f(x)/x(x不等于0),0(x=0),若m>0,求

Answer 1

设x2>x1>=m>0

g(x2)-g(x1)=f(x2)/x2-f(x1)/x1

={[2x2^2+(x2-m)^2]/x2}-{[2x1^2+(x1-m)^2]/x1}

=(3x2-2m+m^2/x2)-(3x1-2m+m^2/x1)

=3(x2-x1)+(m^2/x2-m^2/x1)

=(x2-x1)[(3x1*x2-m^2)/(x1*x2))>0

故g(x)在[m,+无穷）上单调递