Chapter 4.1, Problem 3E

(a)

To determine

To find: Domain and range of the mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{1}{\mathrm{x}+1}\right\}$ .

Answer to Problem 3E

The domain is

  $\mathrm{ℝ}\backslash \{-1\}=(-\mathrm{\infty },-1)\cup (-1,\mathrm{\infty })$

The range is $\begin{array}{l}\mathrm{ℝ}\backslash \left\{0\right\}=(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })\\ \end{array}$

One co domain is $\mathrm{ℝ}$ and other codomain is $(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$ .

Explanation of Solution

Given Information:

The mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{1}{\mathrm{x}+1}\right\}$

Formula Used:

Domain is set of all points where the function is defined.

Codomain is the set containing the set $\mathrm{A}=\{\mathrm{y}:\mathrm{y}=\mathrm{f}(\mathrm{x})\}$

Range is the set $\left\{\mathrm{y}:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\right\}$

Proof:

Consider the given function.

  $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{1}{\mathrm{x}+1}\right\}$

The function $\mathrm{y}=\frac{1}{\mathrm{x}+1}$ is not defined if denominator is zero.

  $\begin{array}{l}\mathrm{x}+1=0\\ \mathrm{x}=-1\end{array}$

The domain is

  $\mathrm{ℝ}\backslash \{-1\}=(-\mathrm{\infty },-1)\cup (-1,\mathrm{\infty })$

Consider the equation.

  $\begin{array}{c}\mathrm{y}=\frac{1}{\mathrm{x}+1}\\ (\mathrm{x}+1)\mathrm{y}=1\\ \mathrm{x}\mathrm{y}+\mathrm{y}=1\\ \mathrm{x}\mathrm{y}=1-\mathrm{y}\\ \mathrm{x}=\frac{1-\mathrm{y}}{\mathrm{y}}\end{array}$

The above is not defined for $\mathrm{y}=0$ .

The range is $\begin{array}{l}\mathrm{ℝ}\backslash \left\{0\right\}=(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })\\ \end{array}$

One co domain is $\mathrm{ℝ}$ and other codomain is $(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$ .

(b)

To determine

To find: Domain and range of the mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}={\mathrm{x}}^2+5\right\}$ .

Answer to Problem 3E

The domain is $\mathrm{ℝ}$.

The range is $\{\mathrm{y}:\mathrm{y}\ge 5\}=[5,\mathrm{\infty })$

One co domain is $\mathrm{ℝ}$ and other codomain is $[5,\mathrm{\infty })$ .

Explanation of Solution

Given Information:

The mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}={\mathrm{x}}^2+5\right\}$

Formula Used:

Domain is set of all points where the function is defined.

Codomain is the set containing the set $\mathrm{A}=\{\mathrm{y}:\mathrm{y}=\mathrm{f}(\mathrm{x})\}$

Range is the set $\left\{\mathrm{y}:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\right\}$

Proof:

Consider the given function.

  $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}={\mathrm{x}}^2+5\right\}$

The function $\mathrm{y}={\mathrm{x}}^2+5$ is defined for all $\mathrm{x}\in \mathrm{ℝ}$ .

The domain is $\mathrm{ℝ}$.

Consider the equation.

  $\mathrm{y}={\mathrm{x}}^2+5\ge 0+5=5\because {\mathrm{x}}^2\ge 0,\forall \mathrm{x}\in \mathrm{ℝ}$

The range is $\{\mathrm{y}:\mathrm{y}\ge 5\}=[5,\mathrm{\infty })$

One co domain is $\mathrm{ℝ}$ and other codomain is $[5,\mathrm{\infty })$ .

(c)

To determine

To find: Domain and range of the mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\tan (\mathrm{x})\right\}$ .

Answer to Problem 3E

Domain is $\mathrm{ℝ}\backslash \left\{\frac{(2\mathrm{n}+1)\mathrm{\pi }}{2}:\mathrm{n}\in \mathrm{\mathbb{Z}}\right\}$ .

Range is $\mathrm{ℝ}$ .

Codomain is $\mathrm{ℝ}$ .

Explanation of Solution

Given Information:

The mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\tan (\mathrm{x})\right\}$

Formula Used:

Domain is set of all points where the function is defined.

Codomain is the set containing the set $\mathrm{A}=\{\mathrm{y}:\mathrm{y}=\mathrm{f}(\mathrm{x})\}$

Range is the set $\left\{\mathrm{y}:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\right\}$

Proof:

Consider the given function.

  $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\tan (\mathrm{x})\right\}$

The function is not defined for $\mathrm{x}=\frac{(2\mathrm{n}+1)\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{\mathbb{Z}}$ .

Domain is $\mathrm{ℝ}\backslash \left\{\frac{(2\mathrm{n}+1)\mathrm{\pi }}{2}:\mathrm{n}\in \mathrm{\mathbb{Z}}\right\}$ .

For every $\mathrm{x}\in \mathrm{ℝ},\exists \mathrm{y}\in \mathrm{ℝ}$ such that $\mathrm{x}={\tan }^{-1}(\mathrm{y})$

Range is $\mathrm{ℝ}$ .

Codomain is $\mathrm{ℝ}$ .

(d)

To determine

To find: Domain and range of the mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}={\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})\right\},{\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})={\{}_{0\text{ otherwise}}^{1\text{ if }\mathrm{x}\in \mathrm{\mathbb{N}}}$ .

Answer to Problem 3E

Domain is $\mathrm{ℝ}$ .

  $\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}=\{0,1\}$

Codomain is $\text{[0,1], and other possaible is range}=\{0,1\}$ .

Explanation of Solution

Given Information:

The mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}={\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})\right\},{\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})={\{}_{0\text{ otherwise}}^{1\text{ if }\mathrm{x}\in \mathrm{\mathbb{N}}}$

Formula Used:

Domain is set of all points where the function is defined.

Codomain is the set containing the set $\mathrm{A}=\{\mathrm{y}:\mathrm{y}=\mathrm{f}(\mathrm{x})\}$

Range is the set $\left\{\mathrm{y}:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\right\}$

Proof:

Consider the given function.

  $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}={\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})\right\},{\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})={\{}_{0\text{ otherwise}}^{1\text{ if }\mathrm{x}\in \mathrm{\mathbb{N}}}$

  $\begin{array}{l}\text{ The function }\mathrm{y}={\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})\text{is defined }\forall \mathrm{x}\in \mathrm{ℝ}\\ \text{ Thus, Domain}=\mathrm{ℝ}\\ \text{For every }\mathrm{x}\in \mathrm{ℝ},\text{ The function }\mathrm{y}={\mathrm{\chi }}_{\mathrm{\mathbb{N}}}(\mathrm{x})\text{ can take only}\\ \text{ possible one two values 0 and 1}\text{.}\\ \Rightarrow \mathrm{R}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}=\{0,1\}\\ \text{ The one possible co-domain is [0,1], and other possaible is range}=\{0,1\}\end{array}$

(e)

To determine

To find: Domain and range of the given mapping.

Explanation of Solution

Given Information:

The mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right\}$

Formula Used:

Domain is set of all points where the function is defined.

Codomain is the set containing the set $\mathrm{A}=\{\mathrm{y}:\mathrm{y}=\mathrm{f}(\mathrm{x})\}$

Range is the set $\left\{\mathrm{y}:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\right\}$

Proof:

  $\begin{array}{l}\text{Given }\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\right\},\\ \text{ The function }\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\text{is defined }\forall \mathrm{x}\in \mathrm{ℝ}\\ \text{ Thus, Domain}=\mathrm{ℝ}\\ \text{For every }\mathrm{y}\in \mathrm{ℝ},\exists \mathrm{x}\in \mathrm{ℝ}\text{ such that }\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}\\ \Rightarrow \text{Range}=\mathrm{ℝ}\\ \text{ The only possible co-domain is }\mathrm{ℝ}.\end{array}$

(f)

To determine

To find: Domain and range of the given mapping.

Explanation of Solution

Given Information:

The mapping $\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{{\mathrm{x}}^2-4}{\mathrm{x}-2}\right\}$

Formula Used:

Domain is set of all points where the function is defined.

Codomain is the set containing the set $\mathrm{A}=\{\mathrm{y}:\mathrm{y}=\mathrm{f}(\mathrm{x})\}$

Range is the set $\left\{\mathrm{y}:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\right\}$

Proof:

  $\begin{array}{l}\text{ Given }\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:\mathrm{y}=\frac{{\mathrm{x}}^2-4}{\mathrm{x}-2}\right\},\\ \text{ The function }\mathrm{y}=\frac{{\mathrm{x}}^2-4}{\mathrm{x}-2}=\frac{(\mathrm{x}-2)(\mathrm{x}+2)}{\mathrm{x}-2}\text{=}(\mathrm{x}+2)\text{is defined }\forall \mathrm{x}\in \mathrm{ℝ}\\ \text{ Thus, Domain}=\mathrm{ℝ}\\ \text{For every }\mathrm{y}\in \mathrm{ℝ},\exists \mathrm{x}=\mathrm{y}-2\in \mathrm{ℝ}\text{ such that }\mathrm{y}=\mathrm{x}+2\\ \Rightarrow \text{Range}=\mathrm{ℝ}\\ \text{ The only possible co-domain is }\mathrm{ℝ}.\end{array}$