Question

若X＜0，则函数f（x）=x²+1/x²-x-1/x的最小值是___？

Answer 1

f(x)=(x+1/x)^2-(x+1/x)-2=t^2-t-2, t<=-2
=(t-1/2)^2-2.25
当t=-2, 即x=-1,得最小值为4.

Answer 2

f(x)=(x^2+1/x^2)-(x+1/x)
f(x)=(x+1/x)^2-(x+1/x)-2
设t=x+1/x x<0 t<=-2
f(t)=t^2-t-2=(t-2)(t+1) (负无穷，1/2）减 (1/2,正无穷）增
t=-2时 取最小
min=4+2-2=4