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Chapter 1: Problem 57

Give an example of a function whose domain is \\{2,5,7\\} and whose range is \\{-2,3,4\\}

Short Answer

Expert verified
The function that has a domain of {2,5,7} and a range of {-2,3,4} is a piecewise function defined as follows: \(f(x) = \begin{cases} -2 & \text{if} \ x=2 \\ 3 & \text{if} \ x=5 \\ 4 & \text{if} \ x=7 \end{cases}\)

Step by step solution

01

Create domain-range pairs

Make a list of domain-range pairs where each element in the domain is paired with an element in the range. To do this, we can match the first element in the domain with the first element in the range, the second with the second, and the third with the third. This gives us the pairs (2,-2), (5,3), and (7,4).
02

Define the function

We can now define our function using these domain-range pairs. We can create a piecewise function that maps each domain element to its corresponding range element. The function will look like this: f(x) = \begin{cases} -2 & \text{if} \ x=2 \\ 3 & \text{if} \ x=5 \\ 4 & \text{if} \ x=7 \end{cases} This function has a domain of {2,5,7} and a range of {-2,3,4}, so it meets the requirements of the exercise.

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Key Concepts

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

Understanding Piecewise Functions
A piecewise function is one that is defined by multiple sub-functions, each applicable to a certain part of the domain. They are incredibly useful for modeling situations where a rule, formula, or expression changes depending on the input value. This is like having different rules for different sections of your domain.
For example, in the solution, the piecewise function is defined as:\[f(x) = \begin{cases} -2 & \text{if} \ x=2 \ 3 & \text{if} \ x=5 \ 4 & \text{if} \ x=7\end{cases}\]
Here, each part of the function deals with a specific value of \(x\). This strategy ensures that each domain value has a clearly defined output or range, allowing for a precise match. Piecewise functions are essential in real-world problems where different conditions apply under different circumstances.
Exploring Domain-Range Pairs
Domain-range pairs are fundamental in understanding how a function operates. The domain is the set of all possible inputs, while the range is the set of corresponding outputs.
These pairs work like a matching game where each element in the domain maps to one in the range. For the exercise above, the pairs are:
  • (2, -2)
  • (5, 3)
  • (7, 4)

This means that when you input 2 into the function, it outputs -2, and so on. Creating these pairs helps in understanding the direct relationship between the domain and range.

Knowing domain-range pairs becomes especially important when dealing with functions where each input requires careful assignment to ensure accuracy.
Mapping Functions Clearly
Mapping functions ensures that each input has a specific and clear output, which is crucial for understanding functional relationships.
In our piecewise example, mapping means linking each specific value in the domain to a value in the range accurately. Think of it as a roadmap that guides each domain element to its correct output in the range. It's like assigning each student a specific locker number at school.

This method guarantees no ambiguity exists in how the inputs are processed. You always know what output is expected for any given input. This clarity makes it easier to predict outcomes and understand the functionality of a system.

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Most popular questions from this chapter

Suppose \(f\) is the function whose domain is the interval \([-2,2],\) with \(f\) defined by the following formula: $$ f(x)=\left\\{\begin{array}{ll} -\frac{x}{3} & \text { if }-2 \leq x<0 \\ 2 x & \text { if } 0 \leq x \leq 2 \end{array}\right. $$ (a) Sketch the graph of \(f\). (b) Explain why the graph of \(f\) shows that \(f\) is not a one-to-one function. (c) Give an explicit example of two distinct numbers \(a\) and \(b\) such that \(f(a)=f(b)\).

(a) True or false: The sum of an even function and an odd function (with the same domain) is an odd function. (b) Explain your answer to part (a). This means that if the answer is "true", then explain why the sum of every even function and every odd function (with the same domain) is an odd function; if the answer is "false", then give an example of an even function \(f\) and an odd function \(g\) (with the same domain) such that \(f+g\) is not an odd function.

For each of the functions \(f\) given in Exercises \(13-\) 22: (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(\boldsymbol{f}^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verify. ing that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\). $$ f(x)=\frac{2 x}{x+3} $$

Give an example of two different functions \(f\) and \(g\), both of which have the set of real numbers as their domain, such that \(f(x)=g(x)\) for every rational number \(x\).

Suppose \(f\) and \(g\) are functions, each of whose domain consists of four numbers, with \(f\) and \(g\) defined by the tables below: $$ \begin{array}{c|c} {x} & {f}({x}) \\ \hline {1} & 4 \\ 2 & 5 \\ 3 & 2 \\ 4 & 3 \end{array} $$ $$ \begin{array}{c|c} x & g(x) \\ \hline 2 & 3 \\ 3 & 2 \\ 4 & 4 \\ 5 & 1 \end{array} $$ Give the table of values for \(g \circ g^{-1}\).
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