Inverse of Strictly Monotone Function
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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 Real 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:
- $x < y \iff \map f x < \map f y$
Hence:
- $\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} < \map {f^{-1} } {\map f y}$
and so:
- $\map {f^{-1} } x < \map {f^{-1} } y \iff x < y$
Similarly, from the definition of strictly decreasing:
- $x < y \iff \map f x > \map f y$
Hence:
- $\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} > \map {f^{-1} } {\map f y}$
and so:
- $\map {f^{-1} } x < \map {f^{-1} } y \iff x > y$
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 12.9$