Proving the equation $C=2x/(1+x^2)^2$ has two positive solutions I want to show the equation $C=\frac{2x}{(1+x^2)^2}$ has two solutions where $C$ is a constant such that $0<C<\frac{1}{2}$. 
My method was to say at $x=1$, $RHS=\frac{1}{2}>C$ and I'm sure you can see my reasoning from there. But is there a smarter way to go about this than just plugging in a number, which may not work for other graphs where there is a tiny interval of the points where the graphs meet?
 A: The intermediate value theorem gives the result, using 


*

*RHS is continuous

*RHS$(1) > C$

*$\lim_{x\to\infty}$ RHS$(x) = 0 < C$

*RHS$(0) = 0 < C$


This seems like a pretty smart way and it works for arbitrary small intervals.
I'm not sure what is meant by "may not work for other graphs where there is a tiny interval of the points where the graphs meet?", but here is yet another method:
You are looking for positive numbers such that $C(1+x^2)^2 = 2x$. Multiplying out yields $$1-\frac{2}{C}x +2x^2 + x^4 = 0 $$ By Descartes' rule of sign this polynomial has either two positive solutions or zero positive solutions.  Now it suffices to find one positive solution.
Another method is to use Sturm sequences for root counting in intervals.
A: Consider the function$$y=\frac{2x}{(1+x^2)^2}$$ Its derivative write $$\frac{dy}{dx}=\frac{2-6 x^2}{\left(x^2+1\right)^3}$$ and it cancels at $x=\pm \frac 1{\sqrt 3}$. At this point the value of $y$ is $\frac{3 \sqrt{3}}{8}$ and the second derivative test would show that this is a maximum. So, for positive $x$, the function increases up to this point and decreases to $0$. So, there are two solutions for any value of $C$ such that $$0\lt C\lt\frac{3 \sqrt{3}}{8}$$ Notice that the upper bound I wrote is larger than $\frac 12$.
