Asymptotic of Inverse Function Suppose we choose a positive constant $c$ and let $f_c(x)=\frac12x^2+cx^{3/2}$. I would like to get an asymptotic estimate for the function $f_c^{-1}(x)$ as $x\rightarrow\infty$. I assume it will be something of the form $f_c^{-1}(x)=\sqrt{2x}+O(g_c(x))$ for some function $g_c(x)$ of order less than $\sqrt x$, but I'm not sure how to get a nice estimate for $g_c(x)$. 
 A: From
$f(x)
=\frac12x^2+cx^{3/2}
=\frac12 x^{2}(1 +2cx^{-1/2})
$,
$\sqrt{f(x)}
=\sqrt{\frac12} x(1 +2cx^{-1/2})^{1/2}
=\sqrt{\frac12} x(1 +cx^{-1/2}+O(1/x))
$.
Therefore
$x 
=\sqrt{2 f(x)}(1 +2cx^{-1/2})^{-1/2}
=\sqrt{2 f(x)}(1 +O(x^{-1/2}))
\approx\sqrt{2 f(x)}
$.
Putting this back in,
$\begin{array}\\
x 
&\approx\sqrt{2 f(x)}(1 +2c\left(\sqrt{2 f(x)}\right)^{-1/2})^{-1/2}\\
&=\sqrt{2 f(x)}(1 +d(f(x))^{-1/4})^{-1/2}
\quad\text{(where } d=c\sqrt{2})\\
&\approx\sqrt{2 f(x)}(1 +(-1/2)d(f(x))^{-1/4}+\frac{(-1/2)(-3/2)}{2}d^2(f(x))^{-1/2}+O(f(x)^{-3/4}))\\
&=\sqrt{2 f(x)} -\frac{d}{\sqrt{2}}(f(x))^{1/4}+\frac{3d^2}{8}+O(f(x)^{-1/4}))\\
&=\sqrt{2 f(x)} -c(f(x))^{1/4}+\frac{3c^2}{4}+O(f(x)^{-1/4}))
\end{array}
$
Anyway,
something like this
should be true.
A: I can propose this method. Let
$$
y = \frac{x^2}{2} + cx^{3/2}.
$$
$y$ is strictly increasing (when $c\ge0$), so inverse function exists. And we have $y\to\infty$ when $x\to\infty$. $x^{3/2}=o(x^2)$, and you start correctly:
$$
x\sim\sqrt{2y}.
$$
Next step:
$$
x\sim\sqrt{2y}(1+z(y)),
$$
where $z(y) = o(1)$. Put back:
$$
y = y(1+z)^2 + c(2y)^{3/4}(1+z)^{3/2},
$$
or
$$
y(2z+z^2) + c(2y)^{3/4}(1+z)^{3/2} = 0.
$$
Since $z=o(1)$, we can write
$$
y\cdot 2z + c(2y)^{3/4}\left(1+\frac32z\right) = 0,
$$
and
$$
z = -\frac{2c}{3c+2(2y)^{1/4}}.
$$
Hence
$$
x\sim \sqrt{2y}\left(1 -\frac{2c}{3c+2(2y)^{1/4}}\right)
$$
A: Consider the equation $$F=\frac12x^2+cx^{3/2}-y$$ and use the first iteration of Newton method starting at $x_0=\sqrt{2y}$. What is obtained is $$x_1=\sqrt{2y}\Big(1-\frac{2 c}{3 c+2 \sqrt[4]{2y} }\Big)$$ which is looking quite close to Michael Galuza's answer.
The same could be done using Halley method but the result would be much more complex.
For illustration purposes, let us use $c=10$, $y=10^{10}$; the approximation leads to $x=137805.$ while the solution is $x=137758.$ 
Edit
We can also do other things such as applying Newton method a few times to update the solution and then expand the result as a Taylor series for infinitely large values of $y$. It seems that the expansion tends to converge to $$x=(2y)^{1/2}-c(2y)^{1/4}+c^2-\frac {7 c^3}8(2y)^{-1/4}+\frac {5 c^4}8(2y)^{-1/2}-\frac{39 c^5}{128}(2y)^{-3/4}+\cdots$$ which, for the example, gives $137758.4700$ which is the solution for $10$ significant figures.
We also could build the simplest Pade approximant at $x=\sqrt{2y}$ and solve. This would lead to $$x=\frac{3 c^2 \sqrt{z}+12 c z^{3/4}+8 z}{15 c^2+20 c \sqrt[4]{z}+8 \sqrt{z}}$$ where $z=2y$.
Edit
The problem looked worth for generalization considering the equation $$F=ax^m+b x^n-y$$ where $m>n>0$, $a>0,b>0$ and infinitely large values of $y$.
Defining $$x=z \left(\frac{y}{a}\right)^{\frac{1}{m}}$$ $$\alpha=\frac{b }{y}\left(\frac{y}{a}\right)^{\frac{n}{m}}$$ the equation reduces to $$F=z^m+\alpha z^n-1$$ Using $z_0=1$, repeating Newton iterations for a few times and expanding the result as Taylor series built at $\alpha=0$ gives the following asymptotics $$z=1-\frac{\alpha }{m}+\frac{\alpha ^2 (-m+2 n+1)}{2 m^2}+\frac{\alpha ^3 \left(-2 m^2+9
   m n+3 m-9 n^2-6 n-1\right)}{6 m^3}+O\left(\alpha ^4\right)$$
