Inverse of Linear Function on Real Numbers
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Theorem
Let $a, b \in \R$ be real numbers such that $a \ne 0$.
Let $f: \R \to \R$ be the real function defined as:
- $\forall x \in \R: \map f x = a x + b$
Then the inverse of $f$ is given by:
- $\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$
Proof
We have that Linear Function on Real Numbers is Bijection.
Let $y = \map f x$.
Then:
\(\ds y\) | \(=\) | \(\ds \map f x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a x + b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac {y - b} a\) |
and so:
- $\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $3$. Mappings: Exercise $4$