Let [tex]$X=\{1, 3, 5\}$[/tex] and [tex]$Y=\{s, t, u, v\}$[/tex]. Define [tex]$f: X \rightarrow Y$[/tex] by the following arrow diagram.

1. Write the domain of [tex]$f$[/tex] and the co-domain of [tex]$f$[/tex].
2. Find [tex]$f(1), f(3)$[/tex], and [tex]$f(5)$[/tex].
3. What is the range of [tex]$f$[/tex]?
4. Is 3 an inverse image of [tex]$s$[/tex]? Is 1 an inverse image of [tex]$u$[/tex]?
5. What is the inverse image of [tex]$s$[/tex]? of [tex]$u$[/tex]? of [tex]$v$[/tex]?
6. Represent [tex]$f$[/tex] as a set of ordered pairs.



Answer :

Let's tackle this problem step-by-step.

### 1. Domain and Co-domain of [tex]\( f \)[/tex]
The domain of a function is the set of all input values for which the function is defined. The co-domain is the set from which the function's output values are chosen.

- Domain: Since [tex]\( f \)[/tex] is defined for [tex]\( X = \{1, 3, 5\} \)[/tex], the domain of [tex]\( f \)[/tex] is [tex]\( X \)[/tex].
[tex]\[ \text{Domain of } f = \{1, 3, 5\} \][/tex]

- Co-domain: Given [tex]\( Y = \{s, t, u, v\} \)[/tex], the co-domain of [tex]\( f \)[/tex] is [tex]\( Y \)[/tex].
[tex]\[ \text{Co-domain of } f = \{s, t, u, v\} \][/tex]

### 2. Find [tex]\( f(1), f(3) \)[/tex], and [tex]\( f(5) \)[/tex]
From the function definition:
- [tex]\( f(1) = 't' \)[/tex]
- [tex]\( f(3) = 's' \)[/tex]
- [tex]\( f(5) = 'v' \)[/tex]

So, we have:
[tex]\[ f(1) = t, \quad f(3) = s, \quad f(5) = v \][/tex]

### 3. Range of [tex]\( f \)[/tex]
The range of a function is the set of all values that the function actually takes.

Given the values:
[tex]\[ f(1) = t, \quad f(3) = s, \quad f(5) = v \][/tex]

Therefore, the range of [tex]\( f \)[/tex] is:
[tex]\[ \text{Range of } f = \{t, s, v\} \][/tex]

### 4. Inverse Image of Given Elements
- 3 as an inverse image of [tex]\( s \)[/tex]: We need to check if [tex]\( 3 \)[/tex] is mapped to [tex]\( s \)[/tex] by [tex]\( f \)[/tex].

Since [tex]\( f(3) = s \)[/tex], [tex]\( 3 \)[/tex] is an inverse image of [tex]\( s \)[/tex].
[tex]\[ 3 \text{ is an inverse image of } s. \][/tex]

- 1 as an inverse image of [tex]\( u \)[/tex]: We need to check if [tex]\( 1 \)[/tex] is mapped to [tex]\( u \)[/tex] by [tex]\( f \)[/tex].

Since [tex]\( f(1) \neq u \)[/tex] ([tex]\( f(1) = t \)[/tex]), [tex]\( 1 \)[/tex] is not an inverse image of [tex]\( u \)[/tex].
[tex]\[ 1 \text{ is not an inverse image of } u. \][/tex]

### 5. Inverse Images of Specific Elements
- Inverse image of [tex]\( s \)[/tex]: We find all [tex]\( x \)[/tex] in [tex]\( X \)[/tex] such that [tex]\( f(x) = s \)[/tex].

Since [tex]\( f(3) = s \)[/tex], the inverse image of [tex]\( s \)[/tex] is:
[tex]\[ \{3\} \][/tex]

- Inverse image of [tex]\( u \)[/tex]: We find all [tex]\( x \)[/tex] in [tex]\( X \)[/tex] such that [tex]\( f(x) = u \)[/tex].

None of the elements in [tex]\( X \)[/tex] map to [tex]\( u \)[/tex], thus:
[tex]\[ \{\} \][/tex]

- Inverse image of [tex]\( v \)[/tex]: We find all [tex]\( x \)[/tex] in [tex]\( X \)[/tex] such that [tex]\( f(x) = v \)[/tex].

Since [tex]\( f(5) = v \)[/tex], the inverse image of [tex]\( v \)[/tex] is:
[tex]\[ \{5\} \][/tex]

### 6. [tex]\( f \)[/tex] as a Set of Ordered Pairs
To represent [tex]\( f \)[/tex] as a set of ordered pairs, we pair each element of [tex]\( X \)[/tex] with its image under [tex]\( f \)[/tex].

The pairs are:
[tex]\[ \{(1, 't'), (3, 's'), (5, 'v')\} \][/tex]

Hence, [tex]\( f \)[/tex] as a set of ordered pairs is:
[tex]\[ f = \{(1, t), (3, s), (5, v)\} \][/tex]

This concludes the step-by-step solution to our problem.