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Chapter 1: Problem 35

Suppose \(f\) is a function whose domain equals \\{2,4,7,8,9\\} and whose range equals \(\\{-3,0,2,6,7\\} .\) Explain why \(f\) is a one-to-one function.

Short Answer

Expert verified
A one-to-one function (or injective function) is a function where each element of the domain corresponds to a unique element in the range. The domain of function f has 5 elements \\(\\{2, 4, 7, 8, 9\\}\\) and its range contains 5 elements \\(\\{-3, 0, 2, 6, 7\\}\\). Since the number of elements in the domain is equal to the number of elements in the range and the function is a total function, each element in the domain has a unique corresponding element in the range. Therefore, we can conclude that the function f is a one-to-one (injective) function.

Step by step solution

01

Define a one-to-one function

A function is one-to-one (injective) if, for every \(x_1\) and \(x_2\) in the domain, \(f(x_1) \neq f(x_2)\) whenever \(x_1 \neq x_2\).
02

Compare the number of elements in the domain and range

First, let's compare the number of elements in the domain and the range. The domain has the elements \\{2, 4, 7, 8, 9\\} which is 5 elements, and the range has the elements \\{-3, 0, 2, 6, 7\\}, also 5 elements.
03

Discuss the relationship between the domain and range

Since the number of elements in the domain is equal to the number of elements in the range, and the function is a total function (which means that every element of the domain is associated with an element of the range), we can infer that each element in the domain has a unique corresponding element in the range.
04

Conclusion

Given that each element of the domain of function f has a unique corresponding element in its range, we can conclude that the function f is a one-to-one (injective) function.

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Most popular questions from this chapter

For Exercises \(33-40,\) assume that \(f\) is the function defined by $$ f(x)=\left\\{\begin{array}{ll} 2 x+9 & \text { if } x<0 \\ 3 x-10 & \text { if } x \geq 0 \end{array}\right. $$ Evaluate \(f(1)\).

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

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 \(I(x)=x\). $$ \begin{aligned} &f(x)=2 x^{2}+5, \text { where the domain of } f \text { equals }\\\ &(0, \infty) \end{aligned} $$

Show that the product of two odd functions (with the same domain) is an even function.

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 \((f \circ g)^{-1}\).
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