设x≠0,则函数y=(x+(1⼀x)^2)-1在x等于多少时,有最小值,最小值为多少?写过程

2025-01-04 22:57:02
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回答1:

解答:
y=(x+(1/x)^2)-1
=x+1/x^2-1

当x>0时,
y=(x+(1/x)^2)-1
=x+1/x^2-1
=x/2+x/2+1/x^2-1
>=3*【(x/2)*(x/2)*1/x^2】-1
=3*1/2*1/2*1-1
=3/4-1
=-1/4
当x/2=1/x^2,即x=2^(1/3)时,y取最小值y(min)=-1/4

当x<0时,
y=(x+(1/x)^2)-1
=x+1/x^2-1是增函数,
当x=负无穷大时,y取最小值负无穷大

解答完毕