/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Give an example to show that the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 46

Give an example to show that the sum of two one-to-one functions is not necessarily a one-to-one function.

Short Answer

Expert verified
Consider two one-to-one functions \(f(x) = x\) and \(g(x) = -x\). Their sum is \(h(x) = f(x) + g(x) = x + (-x) = 0\). The function \(h(x) = 0\) is not one-to-one, as all input values map to the same output value, 0. This example demonstrates that the sum of two one-to-one functions is not necessarily a one-to-one function.

Step by step solution

01

Select two one-to-one functions

Let's consider two simple one-to-one functions: \(f(x) = x\) and \(g(x) = -x\) These two functions are one-to-one because each input values within their domain (all real numbers) corresponds to a unique output value in their range (all real numbers).
02

Find the sum of the two functions

Let \(h(x) = f(x) + g(x)\), which represents the sum of the two one-to-one functions. To find the sum, simply add the functions together: \(h(x) = x + (-x) = 0\)
03

Check if the sum is a one-to-one function

It is clear that the sum function \(h(x) = 0\) is not one-to-one. Regardless of the input value for x, the output value will always be 0. This means that multiple input values within its domain (all real numbers) map to the same output value in the range (0), which violates the definition of a one-to-one function. Thus, by our example, we have shown that the sum of two one-to-one functions is not necessarily a one-to-one function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(f\) and \(g\) are functions, each with domain 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 \(f^{-1} ?\)

Show that the composition of two one-to-one functions is a one-to-one function.

Suppose \(f\) and \(g\) are functions, each with domain 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 \(f^{-1} \circ f\).

Suppose \(g(x)=x^{2}+4\), with the domain of \(g\) being the set of positive numbers. Evaluate \(g^{-1}(7)\).

Suppose \(f\) and \(g\) are both odd functions. Is the composition \(f \circ g\) even, odd, or neither? Explain.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.