Interval of an inverse function I'm struggling with a calculus problem.
I have a surface $\mathcal{S}$ parametrized by
$$ (t,y) \mapsto \left( t, \, y+ \frac{t}{1+y^2}, \, \frac{1}{1+y^2} \right) .$$
My textbook now says: for $t < 8/ \sqrt{27}$ the function $y \mapsto x(t,y) = y+ t/(1+y^2)$ has a smooth inverse, say $y = y(t,x)$.
I tried to apply the Dini theorem, the inverse function theorem and so on, but I don't understand where this $8/\sqrt{27}$ pops out from. Can you help me on this?
If you need more details, don't hesitate asking.
Thank you in advance!
 A: Let $U \subseteq \Bbb{R}^2$. Let $f \in C^\infty(U,\Bbb{R}^2)$. The implicit function theorem states that if the Jacobian $J_f(x)$ is invertible, then $f$ is a local diffeomorphism at $x$. Now our function is 
$$f(t,y) = \bigg(t, y+\frac{t}{1+y^2}\bigg)$$
Your question is to ask why when $U$ equals $\{(t,y)\in \Bbb{R}^2: t > 8/\sqrt{27}\}$, each point in $U$ satisfies the assumptions of ift. The $C^\infty$ assumption is OK, so let's check the $J_f$ assumption:
$D_1f_1(t,y) = 1$, $D_1f_2(t,y) = \frac{1}{1+y^2}$, $D_2f_1(t,y) = 0$ and $D_2f_2(t,y) = 1-\frac{2ty}{(1+y^2)^2}$.
We have $\det(J_f(t,y))= 1-\frac{2ty}{(1+y^2)^2}$. So if we can adjust $t$ such that the following equation
\begin{equation}(1+y^2)^2-2ty= 0\end{equation} 
 has no root, then $J_f$ would be invertible. Looking at this WA result, we see that when $t = \frac{8}{\sqrt{27}}$, the equation above would have real roots and if we make $t$ slightly smaller, all roots of the above equation would be in $\Bbb{C}-\Bbb{R}$. 

Edit: When the equation starts to have real roots, that must be a double root. The other two roots are complex. So the curve $y =g(x)= (1+x^2)^2$ and $y = h(x) = 2tx$ would only touch at one point (i.e. the line is tangent to $(1+x^2)^2$) .
[Sorry for changing variable names!]

So as that equation starts to have real roots at $x= x_0$,
$$g'(x_0) = 2(1+x_0^2)(2x_0) = 2t = h'(x_0)$$
and since that is the point of intersection, we also have
$$g(x_0) = (1+x_0^2)^2 = 2tx_0 = h(x_0)$$
Solving these together gives $x_0 = \frac{1}{\sqrt{3}}$, which agrees with the WA result above, and $t = \frac{8}{\sqrt{27}}$.
Credit: This come from my math major friend

If you have some control engineering background (I am a mechanical engineering undergrad), then you can also have your root locus plot, find the breakin point and then find the corresponding $t$.

(See here if you want to know what that is)
