Question

求曲线y=(x-1)/(x^2+1）的拐点坐标

Answer 1

y ' = 1/(x^2+1) + (x-1) * (-2x) / (x^2+1)^2

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

y '' = (2-2x) / (x^2+1)^2 + (1+2x-x^2) * ( - 4x) / (x^2+1)^3

= 2 ( x^3 - 3x^2 - 3x +1 ) / (x^2+1)^3

y '' = 0 => x^3 - 3x^2 - 3x +1 = 0

即 (x+1) (x^2 -4x+1) =0

=> x1=-1,x2= 2-√3,x3= 2+√3

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