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.

In a neighborhood of $x=-1/4$, $y$ cannot be written as a function of $x$

  • $\begingroup$ Doesn't the IFT also require that $f(x,y)=0$ around that neighborhood? $\endgroup$ – Weierstraß Ramirez Sep 28 '14 at 2:05
  • $\begingroup$ Yes, the assumption is that $(x,y)$ is a point such that $f(x,y)=0$. To be a bit more rigorous, I should have written: Let $P=(x_0,y_0)$ be such that $f(x_0,y_0)=0$. Then there will be an neighborhood of $P$ such that $y=g(x)$ if $$\det(2y+1)\Big|_{y=y_0}\neq 0$$ for which you would obtain $y_0\neq -1/2$ $\endgroup$ – David Peterson Sep 28 '14 at 2:12
  • $\begingroup$ Off course, thanks! $\endgroup$ – Weierstraß Ramirez Sep 28 '14 at 2:15
  • $\begingroup$ There is something I dont get, what happens if $x >-\frac{1}{4}$ then implicit says it should have a solution, but this gives a negative discriminant. Whats wrong ? $\endgroup$ – Rene Schipperus Sep 28 '14 at 2:18
  • $\begingroup$ No, the implicit function theorem says there exists a neighborhood, not any neighborhood. If $x>-1/4$ the $f(x,y)\neq 0$ so $x_0$ from earlier cannot satisfy this.. $\endgroup$ – David Peterson Sep 28 '14 at 2:38

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