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