Implicit function theorem part 2

In summary, "Implicit function theorem part 2" delves into the conditions under which a function defined implicitly can be expressed as an explicit function of its variables. It discusses the necessary criteria, such as the continuity of the function and the non-vanishing of the partial derivatives, which ensure the local existence and uniqueness of the implicit function. The theorem is illustrated through examples and applications, highlighting its significance in solving equations where one variable is defined in terms of others.
  • #1
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Homework Statement
see Post
Relevant Equations
Implicit function theorem
Hi,

I'm not sure if I've solved the problem correctly

Bildschirmfoto 2024-07-06 um 16.56.27.png


In order for the Implicit function theorem to be applied, the following two properties must hold ##F(x_0,z_0)=0## and ##\frac{\partial F(x_0,z_0)}{\partial z} \neq 0##. ##(x_0,z_0)=(1,2)## is a zero and ##\frac{\partial F(x_0,z_0)}{\partial z} =-x^2=-1## so both properties are fulfilled


According to the Implicit function theorem, there now exists a function ##g## for which the following relation ##x=g(z_1)## is valid in the neighborhood ##(z_0 - \epsilon, z_0 + \epsilon)##.


Then the following applies ##F(x,z_1)=F(g(z_1),z_1)##
 
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  • #2
You're doing good! keep going. You could end up with a value of some ##\epsilon>0## in the end, because that's the the problem is asking for. Then, you could show that the statement in the problem is true using the value of ##\epsilon## you found and the implicit function theorem.
 

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