The inverse and implicit function theorems

Section 23.4 The inverse and implicit function theorems

Objectives

Describe continuity in terms of preimages of open sets.
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Prove when a function has a differentiable inverse using the inverse function theorem.
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Establish local solvability of equations using the implicit function theorem.
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Introduction goes here.

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The preimage of a set \(B\subseteq Y\) under a function \(f:X\to Y\) is the set

\begin{equation*} f\inv(B)=\set{x\in X:f(x)\in B}\text{.} \end{equation*}

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A function is continuous if the preimage of every open set is open.
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Inverse function theorem: A continuous function of a single variable \(f:A\to\bb R\) is invertible on its image if and only if it is strictly monotonic. If \(f\) is continuously differentiable on an interval \(I\) and \(f'(x)\neq 0\) for each \(x\in I\text{,}\) then \(f\inv\) is continuously differentiable on \(f(I)\) and

\begin{equation*} (f\inv)'(y)=\dfrac{1}{f'(x)}\text{.} \end{equation*}

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In multiple dimensions, the Jacobian needs to be nonzero, and it guarantees the existence of a unique solution.
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If we drop the assumption that \(f'\) is continuous, \(f\) may not be invertible. Great example is

\begin{equation*} f(x)=x+2x^{2}\sin\tfrac1x\text{.} \end{equation*}

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If \(f'\) is invertible, it maps open sets to open sets (it’s an open map).
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Implicit function theorem: If \(f(x,y)\) is a function that is continuously differentiable in a neighborhood of \((x_{0},y_{0})\text{,}\) and \(\pdv{f}{y}(x_{0},y_{0})\neq 0\text{,}\) then there exists a unique differentiable function \(\phi\) such that \(y_{0}=\phi(x_{0})\) and \(f(x,\phi(x))=0\) in a neighborhood of \(x_{0}\text{.}\) We have:

\begin{equation*} \dv{y}{x}=\phi'(x)=-\frac{\partial f/\partial x}{\partial f/\partial y} \end{equation*}

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