I was asked to determine when the equation $f(x,y)=y^2+y+3x+1=0$ for $y$ in terms of $x$. First the I was asked to provide and answer without using the Implicit Function Theorem (IFT), so I simply analyzed it as a quadratic equation and concluded that the equation had real solutions iff $1-4(1)(3x+1)\geq 0$ which means that $x\geq -1/4$.
The second part of the problem ask you to apply the IFT to determine the exact same thing, in order to compare both results. I know what the IFT states, its just that I'm not sure how using it can help me solve $y$ in terms of $x$.
I' m also asked to compute $\frac{dy}{dx} $, so I did the following: $ \frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}$$=\frac{-3}{2y+1}$. Is this right?
The implicit function theorem gives us:
$$\dfrac{\partial F}{\partial(x,y)} =\left[3 \ \ \ \ 2y+1\right]$$
In any neighborhood of $x$ such that $\det(2y+1)\neq 0$ there exists a function $g(x)$ such that $F(x,y) = F(x,g(x))$. So as long as $y\neq -1/2$.
To get the equivalent answer without the $IFT$, the roots of the quadratic are (where $D$ is the discriminant)
$$y= \dfrac{-1\pm \sqrt{D}}{2}$$
We want to know when $y$ can be written as function of x. And that's when either
$$y = \dfrac{-1+ \sqrt{D}}{2} \ \ \ \ \ or \ \ \ \ \ y = \dfrac{-1- \sqrt{D}}{2}$$
But not both.
However in a neighborhood such that $D=0$ holds, $y$ is $not$ a function of $x$ since it will contain points for which one must switch the definition of $y$ above. Equivalently, there exists an invertible function on some neighborhood such that $y\neq -1/2$ holds.
Consider a neighborhood around $(-1/4,-1/2)$ below.