已知函数f(x)=(x^2-2kx+k^2+1)/x-k

f(x)=(x^2-2kx+k^2+1)/(x-k),有 x-k≠0,即 x≠k,又 x∈（0,+∞）
故 k≤0
又f(x)=(x^2-2kx+k^2+1)/(x-k)=[（x-k）^2+1]/(x-k)=x-k+1/(x-k)
f'(x)=1-1/(x-k)^2,令f'(x)=0,即 (x-k)^2=1,得 x=k±1
此时,f(x)=±2,函数f(x)是双勾函数,
则要 k+1>0,k-1-1,k