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For Problems 7 through 9 determine whether the relationship described is a function. If the relationship is a function, (a) what is the domain? the range? (b) is the function 1 -to- 1 ? $$ \begin{array}{ll} \text { Input } & \text { Output } \\ \hline \sqrt{2} & 0 \\ 2 \sqrt{2} & 0 \\ 3 \sqrt{2} & 0 \\ 4 \sqrt{2} & 0 \end{array} $$

Short Answer

Expert verified
The given relationship is a function with domain \(\{\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}\}\) and range \(\{0\}\). However, it is not a 1-to-1 function.

Step by step solution

01

Determine if it's a function

Review the output for each specific input. In this case, \(\sqrt{2}\), \(2 \sqrt{2}\), \(3 \sqrt{2}\), and \(4 \sqrt{2}\) all have a unique output value which is 0. Therefore, this relationship is a function because every input corresponds to exactly one output.
02

Determine the domain and the range

Next, identify the domain which is the collection of all input values. In this case, that's \(\{\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}\}\). The range is the set of all output values, which here is \(\{0\}\).
03

Determine if the function is 1-to-1

A function is 1-to-1 if no two different elements in the domain correspond to the same element in the range. In this problem, multiple different elements in the domain correspond to the same element 0 in the range. So, it's not a 1-to-1 function.

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Key Concepts

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

Domain and Range
In mathematics, the terms **domain** and **range** are crucial in the study of functions. Think of the domain as the "starting point," or the collection of all possible input values for a function. In simple terms, these are the numbers you plug into the function to see what comes out. For the exercise at hand, the domain is \(\{\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}\}\). This means these are the only specific inputs the function accepts.

On the flip side, the **range** represents all the possible outputs, or the "end points" after calculating the function. In our exercise, the range is \{0\}. This indicates that no matter what input (from the domain) you choose, the output will always be 0. A helpful tip is to remember: all inputs lined up in a simple chart give you the picture of domain and range clearly.
  • The domain is about the inputs, the starting point of your function journey.
  • The range is about outputs, what comes out of the function.
One-to-One Function
**One-to-One** functions have a special characteristic. Imagine if every friend you have has exactly one cell phone number, and no one shares the same number. This is the essence of a one-to-one function, where each input has a unique output, and no two inputs share the same output.

In the exercise, we saw multiple inputs giving the one common output of 0. This tells us the function is not one-to-one because it's like several friends sharing just one cell phone number. For our function to be one-to-one, each input should connect to its own distinct output.
  • A one-to-one function means each input has its own unique output.
  • In the given exercise, multiple inputs share the output of 0, so it's not one-to-one.
Mathematical Functions
A **mathematical function** can be thought of as a special rule that takes every input from the domain and assigns it exactly one output in the range. In a way, it's like a machine that processes numbers based on a set rule. For example, if the rule is "add 3," then the function takes an input and "adds 3" to give the output.

The key idea of a function is the consistency: same input always gives same output. In our exercise, the rule for each input is "output 0," showing how functions operate predictably. Understanding functions includes knowing both the domain and the range, as well as whether it maps inputs uniquely (one-to-one).
  • Functions are predictable rules that connect inputs to outputs.
  • Every input has exactly one output, making the function concept reliable.

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Most popular questions from this chapter

Express the area of a circle as a function of: (a) its diameter, \(d\). (b) its circumference, \(c .\)

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