# Inverse Function (pt1)

Hello:

My 2 questions are underlined below. I have another problem but will first wait for an answer on this before posting the other question.

Problem Statement

Consider the function f:R2->R2 defined by
f(x1,x2) = ( exp (x1-x2) + x$$^{2}_{1}$$x2 + x1(x2-1)$$^{4}$$,
1 + x$$^{2}_{1}$$ + x$$^{4}_{1}$$ + (x1x2)$$^{5}$$ )

Q1- Find Df(x1,x2) at point (1,1) and show that there are open sets (1,1)$$\in$$ U and f(1,1)$$\in$$ V such that f:U->V has an inverse g

Q2- Compute Dg(f(1,1))

Solution

First, I can compute the matrix Df(x) by taking the partial derivatives wrt x1, x2 and then take its determinant evaluated at x= (1,1).
If I evaluate it at (1,1) it comes out to A = [3 0 ; 11 5]

Second, the determinant of A is 15 which is != 0. So an inverse exists at this point.

My question for Q1 above: is it sufficient to prove that if det !=0 then an inverse exists. Is this all the question is askign. What does the question mean when asking to show there are open sets (1,1) $$\in$$ U ....

My question for Q2 above: Dg(f(1,1)) = inv(A) evaluated at (1,1). Is this correct

Thanks

Asif

## Answers and Replies

Q1 is just the inverse function theorem.
Q2 is a special case of the chain rule.

For Q2:

For a chain rule, don't I need a matrix for g. How do I get this?

Is this g matrix, the inverse of f evaluated at (2,4) <-- I get this from f(1,1) = (2,4)
And multiply this by Df(1,1)

Thanks

Asif

Well, you can formulate the chain rule in terms of the jacobian evaluated at p.

J(f○g)(p) = J(g)(f(p))*J(f)(p)

Thanks... yes I think got the Jacobian part.
However, what I am not getting is how do I get 'g'.

BTW, I assume you are saying p=f(1,1)

Thanks

Asif

No, my p in your case is (1,1).

What happens on the left side when g is f's inverse?
What can you then do to solve for J(g)(f(p))?

When g is f's inverse, all I have to do is take inverse of f matrix evaluated at (1,1). This will give me the Jg(f(p)). Is this right?
What I had missed was the Jf(p)

Thanks

Asif

If g is f's inverse, then J(f○g)(p) = Id
So
Id = J(g)(f(p))*J(f)(p)
Then
J^(-1)(f)(p) = J(g)(f(p))