Is there a unique function satisfying $e^{g(x)}+(1+e^x)g(x)^3-x^2=1,\quad\forall x\in\Bbb R$?
Justify your answer.
Note: $g(x)^3=g(x)\cdot g(x)\cdot g(x)$.
My attempt:
I first considered some algebraic manipulations, but I couldn't express $g(x)$ in terms of $x^2, e^x$ or any other elementary function, so I looked up for the Lambert W function, but the expression I have isn't of the form: $$e^{g(x)}+bg(x)+c,\quad b,c\in\Bbb R.$$
I turned to analyzing its behaviour.
$e^{g(x)}+(1+e^x)g(x)^3=1+x^2\ge 1$
Let $h(x)=e^{g(x)}+(1+e^x)g(x)^3=1+x^2,\space h(x)$ is even, differentiable and has a global minimum at $x=0$ and is decreasing on $(-\infty,0)$ and increasing on $[0,+\infty)$.
$e^{g(x)}>0,\space e^x+1>1\space\forall x\in\Bbb R$, so
$$\begin{aligned}e^{g(x)}=1+x^2-(1+e^x)g(x)^3&>0\\ g(x)^3&<\frac{1+x^2}{1+e^x}\\\iff g(x)&<\sqrt[3]{\frac{1+x^2}{1+e^x}}\\\implies\lim_{x\to \color{red}{+}\infty}g(x)&<\lim_{x\to\color{red}{+}\infty}\sqrt{\frac{1+x^2}{1+e^x}}=0\end{aligned}$$
But I couldn't conclude anything for $g(x)$ except that $\lim\limits_{x\to+\infty}e^{g(x)}<1$ and $\lim\limits_{x\to-\infty}\frac{1+x^2}{1+e^x}=\infty$.
I tried to differentiate $h(x)$: $$h'(x)=g'(x)e^{g(x)}+e^xg(x)^3+(1+e^x)3g'(x)g(x)^2,$$ but it doesn't seem fruitful.
Is there another way of determing whether such unique function exists, that I failed to see?
Thank you in advance!
It asks whether there exists a unique $y$ such that $e^{y} + (1+e^{x}) y^{3} = 1+x^{2}$ for each $x$.
Fix $x$, and consider a function $$ H(y):= e^{y} + (1+e^{x})y^{3} - 1-x^{2}. $$ We see that $\frac{dH}{dy} = e^{y} + 3y^{2}(1+e^{x}) >0$ for all $y \in \mathbb{R}$: $H$ is strictly increasing. Furthermore, $\lim_{y \rightarrow \pm \infty} H(y) = \pm \infty$. Indeed, we compare $H(y)$ with $H_{1}(y) := e^{y} - 1-x^{2}$ and $H_{2}(y):=(1+e^{x})y^{3} - 1 - x^{2}$: $H(y) \ge H_{1}(y)$ and $H(y) \le H_{2}(y)$ always. Now observe $\lim_{y \rightarrow +\infty} H_{1}(y) = +\infty$ and $\lim_{y \rightarrow -\infty} H_{2}(y) = -\infty$.
By the continuity of $H$ and the intermediate value theorem, there exists some $y$ such that $H(y) = 0$. By strict monotonicity of $H$, such $y$ is unique. Thus, for each $x \in \mathbb{R}$, there exists a unique value $y=g(x)$ satisfying the condition you desired.