If \(f\) is a one-to-one function that is continuous on its domain, then its inverse function \(f^{-1}\) is also continuous on its domain.
Let \((a,b)\) be the largest open interval included in the domain of \(f.\) If \(f\) is continuous on its domain, then it is continuous on \((a,b) \). Since \(f\) is one-to-one, then \(f\) is either increasing on \((a,b)\) or decreasing on \((a,b)\).
B-4
Suppose \(f\) is increasing on \((a,b)\). Then \(f^{-1}\) is also increasing. Let \(y_{0} = f(x_{0})\). We need to show that \(f^{-1}\) is continuous at \(y_{0}\), given that \(f\) is continuous at \(x_{0}\). The number \(f^{-1}(y_{0}) = x_{0}\) is in the open interval \((a,b)\). Choose \(\varepsilon > 0\) sufficiently small, so that \(f^{-1}(y_{0})-\varepsilon \) and \(f^{-1}(y_{0}) + \varepsilon \) are also in \((a,b)\). Then choose \(\delta \), so that \[ f [ f^{-1}(y_{0})-\varepsilon] <y_{0}-\delta \qquad {\rm and}\ \qquad y_{0}+\delta < f [f^{-1}(y_{0}) + \varepsilon ] \]
Then whenever \(y_{0}-\delta < y < y_{0} + \delta\), we have \[ f [f^{-1}(y_{0})-\varepsilon ] < y < f [f^{-1}(y_{0}) + \varepsilon ] \]
This means whenever \(0 < \vert y-y_{0}\vert < \delta,\) then \[ f^{-1}(y_{0}) - \varepsilon < f^{-1}(y) < f^{-1}(y_{0}) + \varepsilon\quad\ \hbox{or equivalently,}\quad\ \vert f^{-1}(y) - f^{-1}(y_{0}) \vert < \varepsilon \]
That is, \(\lim\limits_{y\rightarrow y_{0}}f^{-1}( y) =f^{-1}( y_{0}) \), so \(f^{-1}\) is continuous at \(y_{0}.\)
The case where \(f\) is decreasing on \((a,b)\) is proved in a similar way.
Let \(y = f(x) \) and \(x = g(y) \) be inverse functions. If \(f\) is differentiable on an open interval containing \(x_{0}\) and if \(f^{\prime} ( x_{0}) \neq 0\), then \(g\) is differentiable at \( y_{0}=f( x_{0}) \) and \[ \dfrac{d}{dy}g( y_{0}) =\dfrac{1}{f^{\prime} ( x_{0}) } \] where the notation \(f^{\prime} (x_{0}) \) means the value of \( f^{\prime} ( x) \) at \(x_{0}\) and the notation \(\dfrac{d}{dy}g( y_{0})\) means the value of \(\dfrac{d}{dy}g(y) \) at \( y_{0}\).
Since \(f\) and \(g\) are inverses of one another, then \( f(x) = y \) if and only if \(x = g(y)\). So, we have the following identity, where \( g(y_{0}) = x_{0}\): \begin{equation*} \dfrac{g(y)-g(y_{0})}{y-y_{0}}=\dfrac{x-x_{0}}{f(x)-f(x_{0})}=\dfrac{1}{\dfrac{ f(x)-f(x_{0})}{x-x_{0}}} \end{equation*}
By the continuity of an inverse function, the continuity of \(f\) at \(x_{0}\) implies the continuity of \(g\) at \(y_{0}\), and \(y\rightarrow y_{0}\) as \( x\rightarrow x_{0}\).
Now take the limits of both sides of the above identity. Since \(f'(x_0)\neq 0\), we have \begin{equation*} g^{\prime} (y_{0})=\lim_{y\rightarrow y_{0}}\dfrac{g(y)-g(y_{0})}{y-y_{0}}=\dfrac{ 1}{\lim\limits_{x\rightarrow x_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}}}=\dfrac{1}{ f^{\prime} (x_{0})} \end{equation*}