Implicit function

[icy wyvern]

I have a problem with this exercise :
f:R^n->R f(x, y) =x^3 +2x^2 y^2 - xy-4x.
I have to check in which points in the form (-1,y) equation f(-1,y)=3 generates implicit function y=phi(x), find derivatives y'(-1) from previous point, check whether it's reversable in the area x_0=-1 (if yes calculate (phi^-1)'(-1,0)), find maximum and minimum of function v->gradient_v f(-1,0) where |v|=1, and prove that gradient f(-1,0) is orthogonal to f^-1(3) (I was told that gradient is always orthogonal to level set)

[runic ocean]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

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3. Feel free to nominate the person for helper of the week in #helper-nominations

[whole briar]

you mean f: R^2 -> R

implicit mapping theorem, at any rate

[icy wyvern]

Yeah, I think it was supposed to be R^2

[icy wyvern]

I don't know how to do the second point

[whole briar]

separate the instructions more clearly, I don't know what you are referring to

[icy wyvern]

It may not be perfect cause I used Google lens to make it translated

[whole briar]

1. Check differentiability

2. apply the derivative formula

$y'(x_0) = -\frac{f_x(x_0,y(x_0))}{f_y(x_0,y(x_0))}$

[nocturne tinsel]

al8628

[whole briar]

by f_x and f_y I mean the partials w.r.t x and y respectively

differentiability is a given since polynomials are smooth

so you really just have to apply 2.

I don't know which version of implicit mapping theorem you are working with, you might need to adjust g(x,y) = f(x,y)-3, but that changes nothing when you take partials, the constant disappears

ofc keep in mind that the IMT gives sufficient conditions for points around which y=y(x) holds

there may be more such points which you will need to check by hand

as for invertibility - if the derivative is strictly positive (or strictly negative) in some neighborhood, then the function is strictly monotone, hence invertible

[icy wyvern]

Would the derivatives be these? Ignore the gradient

[whole briar]

distinguish between the two y-s

mark them differently or something

these y-s i mean

the first one is implicitly defined at (-1,0) and the other one at (-1,-1/2)

[icy wyvern]

Ok, but other than that they are correct?

[whole briar]

yes

[icy wyvern]

What about point c) and d)?

In c I had to check if they are invertible locally in x0=-1. If so calculate (phi) ^-1(y0) where y0=phi(-1).

In d) I have to find max and
min of function v-> gradient_v f(-1,0) where |v|=1

[whole briar]

I commented about invertibility

you should also have a theorem somewhere which tells you how you can find the derivative of an inverse via the derivative of the initial function

[icy wyvern]

Just found the max and min which are sqrt(2) and - 1. Though, I can't find anywhere how to calculate that derivative

[whole briar]

remember that if g is an inverse of f then g'(x) = 1/f'(g(x))

ring a bell?

assuming f is differentiable of course

[whole briar]

can you show that either y is strictly monotone around the corresponding point?

you need that for invertibility

you were able to find max and min which give you the range

[icy wyvern]

I don't remember how to do it. I only remember that for mapping. It was invertible if its Jacobian's determinant wasn't equal 0

But for that it's not square so it wouldn't work

[whole briar]

no need to get that detailed

you already know what derivative of y is

[icy wyvern]

So basically it's just 1/y'(x)?

[whole briar]

don t get ahead of yourself

first check if they are invertible

[icy wyvern]

I already managed to do it, ty

+close