Question

怎样求函数图形的拐点和凹凸区间y=x+x/（x^2-1）

Answer 1

x≠±1

求导:

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

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

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

=(x^4-3x^2)/(x^2-1)^2

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

y'=0,x=0,±√3

{x>√3或x