Inverse of increasing function is increasing

Statement

Suppose $f$ is an increasing function on its domain. Then, $f$ is a one-one function and the inverse function $f^{-1}$ is also an increasing function on its domain (which equals the range of $f$).

Note

Related facts

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Proof

Compatibility with inverse function theorem

This is not a proof but provides an illustration of why the statement is compatible with the inverse function theorem. In particular, the inverse function theorem can be used to furnish a proof of the statement for differentiable functions, with a little massaging to handle the issue of zero derivatives.

The inverse function theorem states that:

$(f^{-1}{)}^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}$

In particular, this tells us that if $f^{\prime }(f^{-1}(x))>0$, then $(f^{-1}{)}^{\prime }(x)>0$. In particular, if $f^{\prime }>0$ everywhere, $(f^{-1}{)}^{\prime }>0$ everywhere.