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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[tex]^{2}_{1}[/tex]x2 + x1(x2-1)[tex]^{4}[/tex],

1 + x[tex]^{2}_{1}[/tex] + x[tex]^{4}_{1}[/tex] + (x1x2)[tex]^{5}[/tex] )

Q1- Find Df(x1,x2) at point (1,1) and show that there are open sets (1,1)[tex]\in[/tex] U and f(1,1)[tex]\in[/tex] 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) [tex]\in[/tex] U ....

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

Thanks

Asif

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# Homework Help: Inverse Function (pt1)

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