Inverse Mapping/Examples/x^2-4x+5

Examples of Inverse Mappings

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: f \paren x = x^2 - 4 x + 5$

Consider the following bijective restrictions of $f$:

\(\ds f_1: \hointl \gets 2\)

\(\to\)

\(\ds \hointr 1 \to\)

\(\ds f_2: \hointr 2 \to\)

\(\to\)

\(\ds \hointr 1 \to\)

The inverse of $f_1$ is:

$\forall y \in \hointr 1 \to: \map {f_1^{-1} } y = 2 - \sqrt {y - 1}$

The inverse of $f_2$ is:

$\forall y \in \hointr 1 \to: \map {f_2^{-1} } y = 2 + \sqrt {y - 1}$

Proof

Let $y = \map f x$.

Then:

\(\ds y\)

\(=\)

\(\ds x^2 - 4 x + 5\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \paren {\paren {x - 2}^2 - 4} + 5\)

\(\ds \)

\(=\)

\(\ds \paren {x - 2}^2 + 1\)

\(\ds \leadsto \ \ \)

\(\ds y - 1\)

\(=\)

\(\ds \paren {x - 2}^2\)

\(\ds \leadsto \ \ \)

\(\ds x - 2\)

\(=\)

\(\ds \pm \sqrt {y - 1}\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds 2 \pm \sqrt {y - 1}\)

$f_1$ is the bijective restriction of $f$ where $x \le 2$.

Hence the negative square root is taken of $\sqrt {y - 1}$, and so:

$\map {f_1^{-1} } y = 2 - \sqrt {y - 1}$

$f_2$ is the bijective restriction of $f$ where $x \le 2$.

Hence the positive square root is taken of $\sqrt {y - 1}$, and so:

$\map {f_2^{-1} } y = 2 + \sqrt {y - 1}$

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $12 \ \text{(iv)}$