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Chapter 4: Problem 13

A relation in \(x\) and \(y\) is given. Determine if the relation defines \(y\) as a one-to-one function of \(x\). $$\begin{array}{|c|c|c|}\hline & A & B \\\\\hline 1 & x & y \\ \hline 2 & 0.6 & 1.8 \\\\\hline 3 & 1 & -1.1 \\ \hline 4 & 0.5 & 1.8 \\\\\hline 5 & 2.4 & 0.7 \\ \hline\end{array}$$

Short Answer

Expert verified
The relation does not define y as a one-to-one function of x.

Step by step solution

01

- Understand One-to-One Function

A function is one-to-one if and only if each element of the range is mapped to by at most one element of the domain. In simpler terms, each value of y must be associated with a unique value of x.
02

- Examine the Given Relation

Review the list of ordered pairs in the table: (0.6, 1.8), (1, -1.1), (0.5, 1.8), (2.4, 0.7).
03

- Check for Repeated y Values

Inspect the y values to make sure no two different x values result in the same y value: In the given pairs, the y-value 1.8 is associated with both x = 0.6 and x = 0.5.
04

- Determine if It's One-to-One

Since the y-value 1.8 corresponds to two different x values (0.6 and 0.5), the relation is not one-to-one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relation
A relation in mathematics often involves two sets: a domain and a range. The domain consists of input values, while the range holds corresponding output values. In simpler terms, it's a collection of ordered pairs (x, y). Each pair shows how an x value from the domain is related to a y value in the range.

In this exercise, we have pairs like (0.6, 1.8) and (1, -1.1). These represent a relation where each x is mapped to a specific y. To decide if a relationship is a function, each x should map to only one y. But just knowing it's a function isn't enough for this exercise. We need to find out if it's a one-to-one function. So, let's continue!
Function
A function is a special type of relation. For any function, every x value in the domain must map to only one y value in the range. In simple terms, two different x values cannot map to the same y value if we are dealing with a function.

In this example, the table shows several x and y pairs: (0.6, 1.8), (1, -1.1), (0.5, 1.8), and (2.4, 0.7). So, do they follow the rule? Each x has only one corresponding y. Yes, this relation is a function. But is it a one-to-one function? We need to break it down further.
Unique Mapping
For a one-to-one function, each y value in the range must map to a unique x value in the domain. In other words, no two x values should have the same y value.

Looking at our pairs again, we see both x = 0.6 and x = 0.5 give y = 1.8. This means these different x values map to the same y value. Consequently, this violates the unique mapping rule. Therefore, this relation is not a one-to-one function because it does not have unique mapping from y back to x.
Domain and Range
The domain of a relation or function includes all possible x values, and the range includes all possible y values. It's essential to specify the domain and range correctly to understand the behavior of the relation or function.

For our table of pairs:
  • The domain is {0.6, 1, 0.5, 2.4}.
  • The range is {1.8, -1.1, 0.7}.
This final understanding helps us see why this relation fails to be one-to-one: because we don't have a unique mapping from the range back to just one domain element. By carefully analyzing the domain and range, we confirm our conclusion about whether a relation is a one-to-one function or not.

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Most popular questions from this chapter

Fluorodeoxyglucose is a derivative of glucose that contains the radionuclide fluorine- \(18\left({ }^{18} \mathrm{~F}\right) .\) A patient is given a sample of this material containing \(300 \mathrm{MBq}\) of \({ }^{18} \mathrm{~F}\) (a megabecquerel is a unit of radioactivity). The patient then undergoes a PET scan (positron emission tomography) to detect areas of metabolic activity indicative of cancer. After \(174 \mathrm{~min}\), one-third of the original dose remains in the body. a. Write a function of the form \(Q(t)=Q_{0} e^{-k t}\) to model the radioactivity level \(Q(t)\) of fluorine- 18 at a time \(t\) minutes after the initial dose. b. What is the half-life of \({ }^{18} \mathrm{~F}\) ?

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \ln 10=\frac{1}{\log e} $$

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(1024=19^{x}+4\)

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