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.
