Question

(1)求函数y=x2-2x+6/x+1的最小值（2）设x属于R且x2+y2/2=1,求根号下1+y2的最大值

Answer 1

(1)y=(x2-2x+6)/x+1

=x-3+9/(x-1)

=(x-1)+9/(x-1) -2

≥2√[(x-1)*9/(x-1)]-2

=6-2

=4

故最小值为4

（2）设x属于R且x2+y2/2=1,

则y^2=2-2x^2

根号下1+y2=√(3-2x^2)

≤√3

故最大值为√3