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Unformatted text preview: y) , yk
x yk
that is, 1 f 1
(y) = Jf f 1 (y)
ek . yk The next exercise shows that dierentiability is not enough for the inverse
function theorem.
42 Exercise 63 Consider the function f : R2 R2 deﬁned by 0
if x = 0,
f1 (x, y ) =
1
x + 2x2 sin x if x = 0,
f2 (x, y ) = y. Prove that f = (f1 , f2 ) is dierentiable in (0, 0) and Jf (0, 0) = 1. Prove that f
is not onetoone in any neighborhood of (0, 0).
The next exercise shows that the existence of a local inverse at every point
does not imply the existence of a global inverse.
Exercise 64 Consider the function f : R2 R2 deﬁned by
f (x, y ) = (ex cos y, ex sin y ) .
Prove that det Jf (x, y ) = 0 for all (x, y ) R2 but that f is not injective. 43...
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This document was uploaded on 03/31/2014.
 Spring '14

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