The function is
y=x/(x^2-9)
The domain is x in (-oo,-3) uu(-3,3) uu(3,+oo)
The first derivative is calculated with the quotient rule
(u/v)'=(u'v-uv')/(v^2)
Here,
u=x, =>, u'=1
v=x^2-9, =>, v'=2x
y'=dy/dx=(1*(x^2-9)-(x*2x))/(x^2-9)^2
=(x^2-9-2x^2)/(x^2-9)^2
=(-9-x^2)/(x^2-9)^2
There are no critical points
y'!=0, AA x in D_y
Let's calculate the second derivative with the quotient rule
u=-9-x^2, =>, u'=-2x
v=(x^2-9)^2, =>, v'=2(x^2-9)*2x
Therefore,
y''=(d^2y)/dx^2=(-2x(x^2-9)^2-(-9-x^2)(4x(x^2-9)))/(x^2-9)^4
=(-2x^3+18x+36x+4x^3)/(x^2-9)^3
=(2x^3+54x)/(x^2-9)^3
=(2x(x^2+27))/(x^2-9)^3
When y''=0, there is a point of inflection.
(2x(x^2+27))/(x^2-9)^3=0
=>, x=0
Let's make a variation chart
color(white)(a)" Interval "color(white)(aaa)(-oo, -3)color(white)(aaa)(-3,0)color(white)(aaa)(0,3)color(white)(aaa)(3,+oo)
color(white)(a)" Sign y'' "color(white)(aaaaaaaa)-color(white)(aaaaaaaaa)+color(white)(aaaaaa)-color(white)(aaaaaa)+
color(white)(a)" y "color(white)(aaaaaaaaaaaaa)nncolor(white)(aaaaaaaaaa)uucolor(white)(aaaaaa)nncolor(white)(aaaaaa)uu
graph{x/(x^2-9) [-10, 10, -5, 5]}
