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Correct typo in single-variable proof |
→Proof for Single-Variable Functions: The previous proof seemed to imply that, on any intervals I and J, the inverse of f was differentiable, in fact, the interval I must be chosen to guarantee that property Tags: Mobile edit Mobile web edit |
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=== Proof for Single-Variable Functions ===
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>
We may without loss of generality assume that <math>f'(x_0) > 0</math>. Given that <math>D</math> is an open set and <math>f'</math> is continuous at <math>x_0</math>, there exists <math>r > 0</math> such that <math>(x_0 - r, x_0 + r) \subseteq D</math> and<math display="block">|f'(x) - f'(x_0)| < \dfrac{f'(x_0)}{2} \qquad \text{for all } |x - x_0| < r.</math>
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In particular,<math display="block">f'(x) > \dfrac{f'(x_0)}{2} >0 \qquad \text{for all } |x - x_0| < r.</math>
This shows that <math>f</math> is strictly increasing for all <math>|x - x_0| < r</math>. Let <math>\delta > 0</math> be such that <math>\delta < r</math>. Then <math>[x - \delta, x + \delta] \subseteq (x_0 - r, x_0 + r)</math>. By the intermediate value theorem, we find that <math>f</math> maps the interval <math>[x - \delta, x + \delta]</math> bijectively onto <math>[f(x - \delta), f(x + \delta)]</math>. Denote by <math>I = (x-\delta, x+\delta)</math> and <math>J = (f(x - \delta),f(x + \delta))</math>. Then <math>f: I \to J</math> is a bijection and the inverse <math>f^{-1}: J \to I</math> exists. The fact that <math>f^{-1}: J \to I</math> is differentiable follows from the differentiability of <math>f</math>. In particular, the result follows from the fact that if <math>f: I \to \R</math> is a strictly monotonic and continuous function that is differentiable at <math>x_0 \in I</math> with <math>f'(x_0) \ne 0</math>, then <math>f^{-1}: f(I) \to \R</math> is differentiable with <math>(f^{-1})'(y_0) = \dfrac{1}{f'(y_0)}</math>, where <math>y_0 = f(x_0)</math> (a standard result in analysis). This completes the proof.
=== A proof using successive approximation ===
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