How is this not a 1-1 function? Using calculus, how would you prove that $$y={{2\,x}\over{\left(x^3+1\right)}}
$$
is not a 1-1 function?
 A: $\lim\limits_{x\to-\infty}\frac{2x}{x^3+1}=0$, $\lim\limits_{x\to-1_-}\frac{2x}{x^3+1}=+\infty$, but eg. $f(1)>0$.
A: Or even more simple, but using the same point as in the answer of user2345215,
$f(x)=1$ is equivalent to
$$
0=x^3-2x+1=(x^2+x-1)(x-1)=((x+0.5)^2-1.25)(x-1)
$$
and from the last term one sees that there are three real solutions.
A: In fact, it is even more trivial:  if $y = 1$, then $2x = x^3 + 1$, or equivalently, $(x-1)(x^2+x-1) = 0$.  The quadratic factor has a positive discriminant, so it admits two distinct real-valued roots, hence the function is not one-to-one, as it has three distinct solutions to $y = 1$.
But, using calculus, we can also compute $$y' = \frac{2(1-2x^3)}{(x^3+1)^2},$$ which obviously has a critical point at $x = 2^{-1/3}$.  The second derivative is $$y'' = \frac{12x^2(x^3-2)}{(x^3+1)^3},$$ which is negative for $x = 2^{-1/3}$, so this critical point is a local extremum, and so the function cannot be one-to-one.
A: There are several good answers, but you asked for one using calculus, so: by setting the derivative to zero, show there's a maximum, and conclude the function's not one-one. 
