Inverse of Strictly Monotone Function

From ProofWiki
Jump to: navigation, search

Theorem

Let $f$ be a real function which is defined on $I \subseteq \R$.

Let $f$ be strictly monotone on $I$.

Let the image of $f$ be $J$.


Then $f$ always has an inverse function $f^{-1}$ and:

if $f$ is strictly increasing then so is $f^{-1}$
if $f$ is strictly decreasing then so is $f^{-1}$.


Proof

The function $f$ is a bijection from Strictly Monotone Function is Bijective.

Hence from Bijection iff Inverse is Bijection, $f^{-1}$ always exists and is also a bijection.


From the definition of strictly increasing:

$x < y \iff f \left({x}\right) < f \left({y}\right)$

Hence:

$f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff f^{-1} \left({f \left({x}\right)}\right) < f^{-1} \left({f \left({y}\right)}\right)$

and so:

$f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff x < y$


Similarly, from the definition of strictly decreasing:

$x < y \iff f \left({x}\right) > f \left({y}\right)$

Hence:

$f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff f^{-1} \left({f \left({x}\right)}\right) > f^{-1} \left({f \left({y}\right)}\right)$

and so:

$f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff x > y$

$\blacksquare$



Sources