若实数x,y满足方程x2+y2-4x+1=0,则y/(x+1)的最大值为_____,最小值为_____.

Question

若实数x,y满足方程x2+y2-4x+1=0,则y/(x+1)的最大值为_____,最小值为_____.

Answer 1

设y/(x+1)=t,
代入原方程得x^2+(tx+t)^2-4x+1=0
==> (1+t^2)x^2+2t^2x-4x+1+t^2=0,
其判别式不小于0,
故(2t^2-4)^2-4(1+t^2)(1+t^2)≥0
4t^4-16t^2+16-4t^4-8t^2-4≥0
-24t^2+12≥0
t^2≤1/2
-√2/2≤t≤√2/2
最大值为__√2/2___,最小值为_-√2/2____.