Show that $f(x) = xe^{x^2}$ is invertible and determine the 6th degree maclaurin polynomial of $f^{-1}(x)$ Show that $f(x) = xe^{x^2}$ is invertible and determine the 6th degree maclaurin polynomial of $f^{-1}(x)$.
I can see that $f$ is invertible since $x$ is strictly increasing and when $x < 0$ $e^{x^2}$ is strictly decreasing and when $x > 0$ $e^{x^2}$ is strictly increasing.  However, I don't know how to find $f^{-1}$? I've tried by the usual approach of setting $xe^{x^2} = y$ and solve for $x$ but I don't get anywhere. Once I've found the inverse the maclaurin polynomial should be straight forward I think.
 A: It is not necessary calculate $f^{-1}$ analytically because the derivative of $f^{-1}$ can be written in function of the derivative of $f$, infact you have that 
$(f^{-1})’(y)=\frac{1}{f’(f^{-1}(y))}$
You can calculate the other derivatives of $f^{-1}$ starting from this relation. 
In your case you have that 
$f(0)=0$ 
So $f^{-1}(0)=0$ that means that 
$(f^{-1})’(0)=\frac{1}{f’(0)}$
For the second derivative, you have that 
$(f^{-1})’’(y)=(\frac{1}{f’(f^{-1}(y))})’=-\frac{1}{(f’(f^{-1}(y))^2} (f’(f^{-1}(y))’ = $
$=-\frac{f’’(f^{-1}(y))}{(f’(f^{-1}(y))^3} $
So
$(f^{-1})’’(0)= -\frac{f’’(0)}{(f’(0))^3}$
You can understand that this is not a good computable method to calculate the Maclaurin expansion because you must do a lot of calculus. In any case you can expand the function $f$ in Maclaurin polynomial of order $6$ and you can calculate the coefficients of $f^{-1}$ if you impose the following condition: 
$f^{-1}(f(x))=x$
A: Since $f'(x)=(1+2x^2)e^{x^2}$ is positive, the $C^{\infty}$-function $f$ is strictly increasing and it follows that it has an inverse $f^{-1}:f(\mathbb{R})=\mathbb{R}\to \mathbb{R}$. Moreover, the Maclaurin expansion of $f$ is
$$f(x)=x(1+x^2+\frac{x^4}{2}+o(x^5))=x+x^3+\frac{x^5}{2}+o(x^6).$$
Note that that $f^{-1}$ is odd (like $f$) and its Maclaurin expansion is
$$f^{-1}(x)=a_1x+a_3x^3+a_5x^5+o(x^6).$$
Then, in a neighbourhood of $x=0$, we have the identity $x=f(f^{-1}(x))$, which implies
$$\begin{align}
x&=(a_1x+a_3x^3+a_5x^5+o(x^6))+(a_1x+a_3x^3+a_5x^5+o(x^6))^3\\
&\quad+\frac{(a_1x+a_3x^3+a_5x^5+o(x^6))^5}{2}+o(x^6)\\
&=a_1x+(a_1^3x^3+3a_1^2a_3x^5)+(a_1^5x^5)
+o(x^6).\end{align}$$
Now compare the terms on both sides and recall that Maclaurin expansion is unique.
Can you take it from here?
A: This case is easily invertible with the help of the Lambert function link
so proceeding accordingly we get from
$$
y = x e^{x^2}\Rightarrow x = \sqrt{\left(\frac{W(2y^2)}{2}\right)}
$$
NOTE
As long as 
$$
x = f(f^{-1}(x))
$$
knowing that
$$
x e^{x^2} = x\left(\sum_{k=0}^{\infty}\frac{x^{2k}}{k!}\right)
$$
assuming $g_n(x) = \sum_{j=0}^n a_j x^{2j+1}$ as a $n$ order approximation for $f^{-1}(x)$ we can calculate this approximation by equating
$$
x = x\left(\sum_{k=0}^{\infty}\frac{\left(\sum_{j=0}^n a_j x^{2j+1}\right)^{2k}}{k!}\right)
$$
expanding powers and equating  we obtain the coefficients for $n=6$
$$
g_6(x) = x-x^3+\frac{5 x^5}{2}-\frac{49 x^7}{6}+\frac{243 x^9}{8}-\frac{14641 x^{11}}{120}+\frac{371293
   x^{13}}{720}
$$
