Content deleted Content added
m Dating maintenance tags: {{Fact}} |
|||
Line 73:
Yet another proof uses [[Newton's method]], which has the advantage of providing an [[effective method|effective version]] of the theorem: bounds on the derivative of the function imply an estimate of the size of the neighborhood on which the function is invertible.<ref name="hubbard_hubbard">{{cite book |first1=John H. |last1=Hubbard |author-link=John H. Hubbard |first2=Barbara Burke |last2=Hubbard|author2-link=Barbara Burke Hubbard |title=Vector Analysis, Linear Algebra, and Differential Forms: A Unified Approach |edition=Matrix |year=2001 }}</ref>
=== Proof for
We want to prove the following: ''Let <math>D \subseteq \R</math> be an open set with <math>x_0 \in D, f: D \to \R</math> a continuously differentiable function defined on <math>D</math>, and suppose that <math>f'(x_0) \ne 0</math>. Then there exists an open interval <math>I</math> with <math>x_0 \in I</math> such that <math>f</math> maps <math>I</math> bijectively onto the open interval <math>J = f(I)</math>, and such that the inverse function <math>f^{-1} : J \to I</math> is continuously differentiable, and for any <math>y \in J</math>, if <math>x \in I</math> is such that <math>f(x) = y</math>, then <math>(f^{-1})'(y) = \dfrac{1}{f'(x)}</math>.''
| |||