Tangents concurent at (1,1/2) 
I got $y'=\frac{1-x^2}{(1+x^2)^2}$..
after tht I got stuck

If $(s,t)=\left(s,\dfrac{s}{1+s^2}\right)$ is a point of tangency of the graph then the slope there is $m=\dfrac{1-s^2}{(1+s^2)^2}$ so the equation of the line of tangency is
$$ y-\dfrac{s}{1+s^2}= \dfrac{1-s^2}{(1+s^2)^2}(x-s)$$
In order for the line of tangency to contain the point $\left(1,\dfrac{1}{2}\right)$ the following equation must be satisfied:
$$ \frac{1}{2}-\dfrac{s}{1+s^2}= \dfrac{1-s^2}{(1+s^2)^2}(1-s) $$
which simplifies to the polynomial equation
$$ s^4-4s^3+4s^2-1=0 $$
which factors
$$ (s-1)^2(s^2-2s-1)=0 $$
yielding the solutions $1\pm\sqrt{2}$ and the double solution $s=1$
So the points of tangency are


*

*$\left(1,\dfrac{1}{2}\right)$ with slope $m=0$

*$\left(1+\sqrt{2},\dfrac{\sqrt{2}}{4}\right)$ with slope $m=1-\dfrac{\sqrt{2}}{4}$

*$\left(1-\sqrt{2},-\dfrac{\sqrt{2}}{4}\right)$ with slope $m=1+\dfrac{\sqrt{2}}{4}$


The three tangent lines have equations


*

*$y=\dfrac{1}{2}$

*$y=\left(\dfrac{1-\sqrt{2}}{4}\right)x+\dfrac{1+\sqrt{2}}{4}$

*$y=\left(\dfrac{1+\sqrt{2}}{4}\right)x+\dfrac{1-\sqrt{2}}{4}$

