Problem 127
Question
Give an example of a relation with the following characteristics: The relation is a function containing two ordered pairs. Reversing the components in each ordered pair results in a relation that is not a function.
Step-by-Step Solution
Verified Answer
A function that satisfies the conditions given by the exercise is \( f = \{(1,2), (3,4)\} \). When we reverse the order of the pairs, \( R = \{(2,1), (4,3)\} \), which is not a function as no element in the domain maps to more than one element in its codomain.
1Step 1: Construct a function
First, we must construct a function that contains two ordered pairs. An example of such a function would be \( f = \{(1,2), (3,4)\} \). The function \( f \) is defined so that it maps 1 to 2 and 3 to 4.
2Step 2: Reverse the order of the pairs
Next, we reverse the order of the pairs in our function \( f \) to create a new relation \( R \). This means \( R = \{(2,1), (4,3)\} \).
3Step 3: Check if the relation is still a function
Finally, we verify whether the relation \( R \) is still a function or not, according to the definition of a function. Since we only reversed the existing pairs without creating duplicate mappings for elements in the domain, the relation \( R \) is not a function, fulfilling the conditions set by the exercise.
Other exercises in this chapter
Problem 124
will help you prepare for the material covered in the next section. $$ \text { Solve for } y: 3 x+2 y-4=0 $$
View solution Problem 126
If \(f(x)=3 x+7,\) find \(\frac{f(a+h)-f(a)}{h}\)
View solution Problem 128
If \(f(x+y)=f(x)+f(y)\) and \(f(1)=3,\) find \(f(2), f(3)\) and \(f(4) .\) Is \(f(x+y)=f(x)+f(y)\) for all functions?
View solution Problem 129
Exercises \(129-131\) will help you prepare for the material covered in the next section. The function \(C(t)=20+0.40(t-60)\) describes the monthly cost, \(C(t)
View solution