y={1+x}/{1+x^2}
By Quotient Rule,
y'={1cdot(1+x^2)-(1+x)cdot2x}/{(1+x^2)^2}
={1-2x-x^2}/{(1+x^2)^2}
By Quotient Rule,
y''={(-2-2x)cdot(1+x^2)^2-(1-2x-x^2)cdot2(1+x^2)(2x)}/{(1+x^2)^4}
={2(1+x^2)(-1-x^2-x-x^3-2x+4x^2+2x^3)}/{(1+x^2)^4}
={2(x^3+3x^2-3x-1)}/{(1+x^2)^3}
={2(x-1)(x^2+4x+1)}/{(1+x^2)^3}
By setting y''=0,
{(x-1=0 Rightarrow x=1),(x^2+4x+1=0 Rightarrow x=-2pm sqrt{3}):}
Using the x-values found above to split (-infty,infty) into intervals
(-infty,-2-sqrt{3}),(-2-sqrt{3},-2+sqrt{3}),(-2+sqrt{3},1), " and " (1,infty).
Using sample points: x=-4,-2,0, and 2 for the intevals, respectively,
y''(-4)<0, y''(-2)>0, y''(0)<0, and y''(2)>0,
which indicate concavity changes at each point; therefore, there are inflection points at x=1, -2pmsqrt{3}.
The inflection points are
(1,y(1))=(1,1),
(-2-sqrt{3},y(-2-sqrt{3}))=(-2-sqrt{3},{1-sqrt{3
}}/4),
and
(-2+sqrt{3},y(-2+sqrt{3}))=(-2+sqrt{3},{1+sqrt{3}}/4).
I hope that this was helpful.
