Implicit function theorem find neighbourhood of the point

[gothloli]

Homework Statement

F(x,y) = y² - x^4. At the point a = (0.5, 0.25) the implicit function theorem holds. Find the largest r1neighbourhood of a s.t $\frac{\partial F(x,y)}{\partial y}$ >0. Find the largest possible r0 > 0 so that for all x, $\left | x -a \right |$ < r0 implies F(x, 0.25 - r1) < 0 and F(x, 0.25 + r1) >0

Homework Equations

implicit function theorem

The Attempt at a Solution

dyF(x,y) = 2y>0 means y>0, since only condition is that y>0 and 0.25 - r1 < y < 0.25 + r1, r1 = 0.25. But that wouldn't make sense since then F(x, 0.25 - 0.25) = 0 which wouldn't follow the implicit function theorem. My question is how can you find the largest possible values since r0 and r1 can be anything > 0.

 

[mighty2000]

First of all, I would like to clarify that the implicit function theorem is used to find a function that represents a given implicit equation, not to find the largest values of r0 and r1.

In order to find the largest possible values of r0 and r1, we need to analyze the given function F(x,y) = y^2 - x^4 at the point a = (0.5, 0.25).

To satisfy the condition \frac{\partial F(x,y)}{\partial y} > 0, we can see that y^2 - x^4 should be increasing in the y direction. This means that the slope of the function in the y direction should be positive. In other words, the derivative of F(x,y) with respect to y should be positive.

Taking the partial derivative of F(x,y) with respect to y, we get:
\frac{\partial F(x,y)}{\partial y} = 2y > 0
This implies that y > 0.

Now, let's consider the condition \left | x -a \right | < r0 implies F(x, 0.25 - r1) < 0 and F(x, 0.25 + r1) >0.

Since we have already established that y > 0, we can simplify the condition to \left | x -a \right | < r0 implies F(x, 0.25 + r1) >0.

To satisfy this condition, we need to find the largest possible value of r0 such that for all x, the value of F(x, 0.25 + r1) is positive.

To do this, we can first find the minimum value of F(x, 0.25 + r1) at x = 0.5.
F(0.5, 0.25 + r1) = (0.25 + r1)^2 - 0.5^4
= 0.0625 + 0.5r1 + r1^2 - 0.0625
= 0.5r1 + r1^2

Since r1 > 0, r1^2 will always be positive. Therefore, the minimum value of F(0.5, 0.25 + r1) will be when r1 = 0, which gives us a