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The question is:

Let: $$g(x,y)=(e^x+1)y^2+2(e^{x^2}-e^{2x-1})y+(e^{-x^2}-1)$$ Then, show that there exists a constant $\bar{y}>0$, such that for any fixed $y \in [0,\bar{y}]$, the equation $g(x,y) = 0$ admits a solution $x(y)$.

This question is with the background of an exercise about Rolle's Theorem. However, it is still possible that the solution isn't unique.

The problem is I have no idea about how to apply it. I have considered first integrate $g(x, y)$ with respect to $x$, then apply the Rolle's Theorem to the point $0$ and $\bar{y}$ in the integrated function but I failed because I can't integrate $e^{x^2}$ and I failed to find an appropriate $\bar{y}$.

Any help is appreciated! Many thanks.

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It is obvious that for y>0 :$$\lim_{x\to\infty} g(x,y)=+\infty$$ Then we just need to find a constant $\overline{y}$ such that, for $y\in]0,\overline{y}]$ , the inequality : $$g(x,y)<0$$ admits a solution x

Let's take $x=1$, we have $g(1,y)=(e+1)y^2+\frac{1}{e}-1$, so, if we take $\overline{y}=\sqrt{\frac{1-\frac{1}{e}}{e+1}}$ then for every $y\in]0,\overline{y}]$ : $$g(1,y)<0$$ Thus we can apply the role theorem to the function $x\mapsto g(x,y)$ which gives the existence of the solution x(y) for every $y\in]0,\overline{y}]$ Finally, for $y=0$ , the solution is $x(y)=0$ .

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