Question

How do you find the inflection point of `(x+1)/(x^(2)+1)`?

Answer 1

We get the Points
`P_1(1;1),P_2(-2-sqrt(3),1/4*(1-sqrt(3))),P_3(-2+sqrt(3),1/4(1+sqrt(3)))`

Explanation:

By the Quotient rule we get
`f'(x)=(-x^2-2x+1)/(x^2+1)^2`
`f''(x)=(2x^3+6x^2-6x+2)/(x^2+1)^3`
so we have to solve
`2(x^3+3x^2-3x-1)=0`
This is
`(x-1)(x^2+4x+1)=0`
Solutions are
`x_1=1`
`x_2=-2-sqrt(3)`
`x_3=-2+sqrt(3)`

Answer 2

`x_1=1`
`x_2=-2-sqrt(3)`
`x_3=-2+sqrt(3)`

Explanation:

For a point of inflection `g`:
`f''(g)=0`
`f'''(g)ne0`

`f(x)=(x+1)/(x^2+1)`

Using the Quotient rule:
`f'(x)=(1*(x^2+1)-(x+1)(2x))/(x^2+1)^2`
`=(x^2+1-2x^2-2x)/(x^2+1)^2`
`=(-x^2-2x+1)/(x^2+1)^2`
`=-((x+1)^2-2)/(x^2+1)^2`

Again, using the Quotient rule:
`f''(x)=-(2*(x+1)*(x^2+1)^2-((x+1)^2-2)(2*2x(x^2+1)))/(x^2+1)^4`
`=-((2x+2)* (x^2+1)^2-4x(x+1)^2(x^2+1)-4x(x^2+1))/(x^2+1)^4`
`=-((2x+2)*(x^2+1)-4x(x+1)^2-4x)/(x^2+1)^3`

`f''(x)=0`
`0=-((2x+2) * (x^2+1)-4x(x+1)^2-4x)/(x^2+1)^3|* (x^2+1)^3`
`0=-(2x+2) * (x^2+1)-4x(x+1)^2-4x`
`0=-2x^3+2x-2x^2-2+4x^3+8x^2-4x-4x`
`0=2x^3+6x^2-6x-2`

`x_1=1`
`x_2=-2-sqrt(3)`
`x_3=sqrt(3)-2`

Again, using the Quotient rule:
`f'''(x)=-(6(x^4+4x^3+4x^2-4x-1))/(x^2+1)^4`
`f'''(1)=-3/2`
`f'''(-2-sqrt(3))=3/32*(4*sqrt(3)-7)~~-0.0067`
`f'''(-2+sqrt(3))=-3/32*(4*sqrt(3)+7)~~-1.31`