Given the following function:
$$g(x,y) = (e^x+1)y^2+2(e^{x^2}-e^{2x-1})y+(e^{-x^2}-1) $$
I want to show that there exists a $\bar{y} \in \mathbb{R}$ such that $\bar{y} > 0$ and for a fixed $y \in [0,\bar{y}]$ the equation $g(x,y) = 0$ amits a solution $x(y).$
Any ideas on how to proceed?? I can't come up with a way to approach this. Is Rolle's theorem applicable? I can see that at $y =0, g(x,y) = 0 \Rightarrow x =0$. So to apply Rolles's theorem, need another value of $y$ for which $g(x,y) = 0$ gives a similar answer.
Thanks.
Suppose $y > 0$. Notice that $$g(0, y) = 2y^2 + 2(1 - e^{-1}) y$$ is positive for any $y$. On the other hand $$g(1, y) = (e + 1) y^2 + e^{-1} - 1$$ could be negative. In fact,
$$(e + 1) y^2 + e^{-1} - 1 < 0 \quad\Longleftrightarrow\quad y < \sqrt {\frac {1 - e^{-1}} {1 + e}} \approx 0.41$$
Let $\bar y = \sqrt {\frac {1 - e^{-1}} {1 + e}}$. Since for any fixed $y \in [0, \bar y]$ the function $g(x, y)$ is continuous and takes values of opposite sign in $0$ and $1$, by the intermediate value theorem the equation $g(x, y) = 0$ has at least one solution $x(y) \in (0, 1)$.