THEOREM Continuity of the Inverse Function

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.

Proof

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 $(a,b)$.

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.

THEOREM Derivative of the Inverse Function

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 $y_{0}$.

Proof

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 $$\begin{equation*} g^{\prime} (y_{0})=\lim_{y\rightarrow y_{0}}\dfrac{g(y)-g(y_{0})}{y-y_{0}}=\dfrac{ 1}{\lim\limits_{x\rightarrow x_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}}}=\dfrac{1}{ f^{\prime} (x_{0})} \end{equation*}$$