/*! 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 25 For the following exercises, det... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 25

For the following exercises, determine whether the relation represents \(y\) as a function of \(x .\) $$y^{2}=x^{2}$$

Short Answer

Expert verified
The relation does not represent \(y\) as a function of \(x\).

Step by step solution

01

Understand the Definition of a Function

A function is a relation where every input (usually denoted as \(x\)) is mapped to exactly one output (usually denoted as \(y\)).
02

Rearrange the Given Equation

Consider the relation \(y^{2} = x^{2}\). Rearrange it to express \(y\) in terms of \(x\). You can rewrite it as \(y = \pm x\).
03

Analyze the Mapping from x to y

For each value of \(x\), there are two possible values for \(y\): \(y = x\) and \(y = -x\).
04

Determine if Multiple Outputs Exist

Since each \(x\) corresponds to two possible \(y\) values, this means that \(y\) is not uniquely determined for every \(x\).
05

Conclude Whether y is a Function of x

Since \(y\) can have more than one value for each \(x\), the relation \(y^2 = x^2\) does not represent \(y\) as a function of \(x\).

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!

Key Concepts

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

Relation
In mathematics, a relation describes a connection or association between two sets of numbers or elements. For example, if you have a set of inputs and a set of outputs, a relation tells you how they are related. In simple terms, it's like a rule that links each input to its outputs.
Consider the relation given by the equation \( y^2 = x^2 \). This equation links the variables \( x \) and \( y \).
However, just because there is a relation doesn't automatically mean it's a special one called a function. The word 'relation' itself doesn't imply uniqueness or exclusivity of pairing between inputs and outputs.
Input-output Mapping
Input-output mapping refers to how each element from the input set is paired with elements from the output set in a relation. Imagine each \( x \) value as an input you plug in, and the corresponding \( y \) values as the outputs you get.
For the equation \( y^2 = x^2 \), if you try different numbers for \( x \), you'll notice that each input can lead to different outputs.
  • For example, if \( x = 3 \), \( y = 3 \) or \( y = -3 \).
  • If \( x = -2 \), \( y = 2 \) or \( y = -2 \).
These examples show that each input results in two possible outputs. However, for a function, each input should map to exactly one output.
Uniqueness of Outputs
Uniqueness of outputs is a crucial part of defining a function. In a function, every input must relate to one and only one output. This is like saying each key (input) opens exactly one lock (output).
In the equation \( y^2 = x^2 \), for each \( x \) value, there are often two corresponding \( y \) outputs. This breaks the rule of uniqueness, which indicates that our relation isn't a function.
If \( x = 4 \), then \( y \) could be 4 or -4. Such ambiguity in outputs means that the criterion of uniqueness is not met, solidifying the point that the relation is not a function.
Functions versus Non-Functions
Understanding the difference between functions and non-functions is essential in grasping how relations work. A function is a particular type of relation where every single input has one distinct output. This is often represented by an arrow pointing from each input to a single output.
  • If a relation gives multiple outputs for a single input, like our \( y^2 = x^2 \), it's not a function.
  • Consider a function such as \( y = x + 1 \). Here, each \( x \) has one unique \( y \), making it a perfectly valid function.
Understanding these distinctions helps clarify when a relation satisfies the conditions to be classified as a function.

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

For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. $$ k(x)=(x-2)^{3}-1 $$

For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. $$ f(t)=(t+1)^{2}-3 $$

For the following exercises, use the function values for \(f\) and \(g\) shown in Table 1.23 to evaluate each expression. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & {0} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} \\ \hline f(x) & {7} & {6} & {5} & {8} & {4} & {0} & {2} & {1} & {9} & {3} \\ \hline g(x) & {9} & {5} & {6} & {2} & {1} & {8} & {7} & {3} & {4} & {0} \\ \hline\end{array} $$ $$ f(g(8)) $$

For the following exercises, determine the interval(s) on which the function is increasing and decreasing. $$ f(x)=4(x+1)^{2}-5 $$

For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\) $$ h(x)=\left(\frac{8+x^{3}}{8-x^{3}}\right)^{4} $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.