# Preimage of Subset under Mapping/Examples

## Examples of Preimage of Subset under Mapping

### Subset of Image of Square Root Function

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

$\forall x \in \R: \map f x = x^2$

Let $A \subseteq \R$ be defined as:

$A := \closedint 4 9 = \set {x \in \R: 4 \le x \le 9}$

Then the preimage of $A$ under $f$ is:

$f^{-1} \sqbrk A = \closedint {-3} {-2} \cup \closedint 2 3$

Let $B \subseteq \R$ be defined as:

$B := \closedint {-9} {-4} = \set {x \in \R: -9 \le x \le -4}$

Then the preimage of $B$ under $f$ is:

$f^{-1} \sqbrk B = \O$

### Identity Function with Discontinuity

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

$\forall x \in \R: \map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$

Let the following subsets of $\R$ be defined:

 $\ds A$ $=$ $\ds \openint \gets 0$ $\ds B$ $=$ $\ds \openint 0 \to$ $\ds C$ $=$ $\ds \openint 0 1$ $\ds D$ $=$ $\ds \openint {-1} 1$ $\ds E$ $=$ $\ds \openint {-2} {-1}$ $\ds F$ $=$ $\ds \openint {\dfrac 1 2} 2$

Then the preimages under $f$ of these sets is:

$f^{-1} \sqbrk A = \set {x \in \R: \map f x < 0} = \openint \gets 0$
$f^{-1} \sqbrk B = \set {x \in \R: \map f x > 0} = \openint 0 \to$
$f^{-1} \sqbrk C = \set {x \in \R: 0 < \map f x < 1} = \openint 0 1$
$f^{-1} \sqbrk D = \set {x \in \R: -1 < \map f x < 1} = \openint {-1} 0 \cup \openint 0 1$
$f^{-1} \sqbrk {E \cup F} = \openint {-2} {-1} \cup \openint {\dfrac 1 2} 2 \cup \set 0$

### Preimage of $\closedint {-2} 0$ under $\map f x = x^2 - x - 2$

Let $f: \R \to \R$ be the mapping defined as:

$\forall x \in \R: \map f x = x^2 - x - 2$

The preimage of the closed interval $\closedint {-2} 0$ is:

$f^{-1} \closedint {-2} 0 = \closedint {-1} 0 \cup \closedint 1 2$

### Preimage of $\set 0$ under $\map f x = x^2 - x - 2$

Let $f: \R \to \R$ be the mapping defined as:

$\forall x \in \R: \map f x = x^2 - x - 2$

The preimage of the singleton $\set 0$ is:

$f^{-1} \sqbrk {\set 0} = \set {-1, 2}$

which is the set of roots of $f$.

### Preimage of $\closedint {-5} {-4}$ under $\map f x = x^2 - x - 2$

Let $f: \R \to \R$ be the mapping defined as:

$\forall x \in \R: \map f x = x^2 - x - 2$

The preimage of the closed interval $\closedint {-5} {-4}$ is:

$f^{-1} \closedint {-2} 0 = \O$

the Empty Set

### Preimages of $\map f {x, y} = x y$

Yellow: $f^{-1} \sqbrk S$, orange: $f^{-1} \sqbrk T$ (boundaries not included)

Let $f: \R^2 \to \R$ be the real function of $2$ variables defined as:

$\forall \tuple {x, y} \in \R^2: \map f {x, y} = x y$

Let the following subsets of $\R^2$ be defined:

 $\ds S$ $=$ $\ds f^{-1} \sqbrk {\openint 1 \to}$ $\ds T$ $=$ $\ds f^{-1} \sqbrk {\openint 0 1}$

Then:

 $\ds S$ $=$ $\ds \set {\tuple {x, y} \in \R^2: x y > 1}$ $\ds T$ $=$ $\ds \set {\tuple {x, y} \in \R^2: 0 < x y < 1}$

### Preimages of $\map g {x, y} = \tuple {x^2 + y^2, x y}$

Yellow: $g^{-1} \sqbrk {\openint 0 3 \times \openint 0 1}$ (boundary not included)

Let $g: \R^2 \to \R^2$ be the mapping defined as:

$\forall \tuple {x, y} \in \R^2: \map g {x, y} = \tuple {x^2 + y^2, x y}$

Let the following subset of $\R^2$ be defined:

$S = g^{-1} \sqbrk {\openint 0 3 \times \openint 0 1}$

Then:

$S = \set {\tuple {x, y} \in \R^2: x y < 1} \cap \set {\tuple {x, y} \in \R^2: 0 < x^2 + y^2 < 3}$