# how to do calculus question?

let f and g be c^1 functions from r^3-r.show that the mapping F(f,g,f+g):r^3-r^3,does not have a differentiable local inverse anywhere.

please pf more clearly! step by step

### 1 Answer

- mcbengtLv 77 years agoFavorite Answer
Let's compute the total derivative (or Jacobian or whatever you want to call it) of f, Df at an arbitrary point. To be explicit: we have

F(x,y,z) = (f(x,y,z), g(x,y,z), f(x,y,z) + g(x,y,z))

for all (x,y,z) in R^3. Letting h_j denote the partial derivative of a function h with respect to its jth variable, the Jacobian matrix of F is, by definition

DF(x,y,z) = the 3x3 matrix with whose jth row's entries are f_j(x,y,z), g_j(x,y,z), and f_j(x,y,z) + g_j(x,y,z).

What you might notice about DF is that no matter what (x,y,z) happens to be, the third column of DF(x,y,z) is the sum of the first two columns of DF(x,y,Z). This means that no matter what (x,y,z) is, the matrix DF(x,y,z) is not invertible (if you just take its determinant by hand, using the 3x3 determinant formula, you will also see that).

This tells you that F cannot have a differentiable local inverse anywhere, because if F does have such a thing at some point (a,b,c), you could use it to prove that DF(a,b,c) was invertible. [In detail: if there is a differentiable function G such that G(F(x,y,z)) = (x,y,z) for all (X,y,z) near (a,b,c), then differentiating both sides of this equation and using the chain rule, you deduce that the matrix product of DG(F(a,b,c)) and DF(a,b,c) is equal to the identity matrix. Which would tell you that DF(a,b,c) is invertible.]

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