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#