/*! 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 61 Determine whether the equation d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 61

Determine whether the equation defines \(y\) as a function of \(x .\) $$x+y^{2}=9$$

Short Answer

Expert verified
The equation does not define \(y\) as a function of \(x\).

Step by step solution

01

Understand the Equation

We are given the equation \(x + y^2 = 9\), which involves two variables: \(x\) and \(y\). Our aim is to determine if \(y\) can be explicitly defined as a function of \(x\). In a function, for each \(x\) value, there should be only one corresponding \(y\) value.
02

Rewrite the Equation

Start by isolating \(y^2\) in the equation. We can do this by subtracting \(x\) from both sides:\[ y^2 = 9 - x \]
03

Solve for y

To solve for \(y\), take the square root of both sides of the equation:\( y = \pm \sqrt{9 - x} \). This indicates that for a given \(x\), \(y\) can have two possible values: \(\sqrt{9 - x}\) or \(-\sqrt{9 - x}\).
04

Determine Function Conditions

Since \(yeq f(x)\) results in two possibilities for each value of \(x\), \(y\) cannot be written as a function of \(x\). A function assigns exactly one output for each input, while here we have two possible outputs \(\pm\sqrt{9 - x}\).

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.

Solving Equations
Solving equations is all about finding the value of a variable that makes the equation true. In the given example, we work with the equation \(x + y^2 = 9\). The goal is to express one variable in terms of the other. To do this, we start by rearranging the equation to isolate one variable. Here, we isolate \(y^2\) by subtracting \(x\) from both sides:
  • \(y^2 = 9 - x\)
This step involves understanding how the equation works and using basic algebraic operations, such as addition, subtraction, multiplication, and division, to simplify it. Rearranging equations is a key skill in algebra. It allows you to see the relationships between variables more clearly.
Square Root Function
The square root function is pivotal in solving quadratic equations, like the one we derived: \(y^2 = 9 - x\). Taking the square root involves finding a number which, when multiplied by itself, gives the original number. When solving for \(y\), we take the square root of both sides:
  • \(y = \pm \sqrt{9 - x}\)
Here, the plus-minus sign (\(\pm\)) indicates that there are two possible solutions when taking the square root of a number. This is because the square of both a positive and negative number yields the same result. Thus, \(\sqrt{4}\) can be \(2\) or \(-2\). It is crucial to include both possibilities when solving squares, as it reflects the true essence of square root operations.
Multivalued Solutions
In some equations, solving for a variable leads to multivalued solutions. This occurs when a single input value can produce multiple output values. In the equation we investigated, \(y = \pm \sqrt{9 - x}\) implies a multivalued situation:
  • For any given \(x\), \(y\) could be either \(\sqrt{9-x}\) or \(-\sqrt{9-x}\).
A function, by definition, should map each input to a single output. However, since our solution produces two potential values for \(y\) for each \(x\), it does not satisfy the requirements of a function. Understanding multivalued solutions helps us appreciate why certain equations cannot be classified as functions. It also highlights the need for functions to have a one-to-one relationship between inputs and outputs.

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

Express the function (or rule) in words. $$h(x)=x^{2}+2$$

Find the domain of the function. $$f(x)=\frac{1}{x-3}$$

Evaluate the piece wise defined function at the indicated values. $$\begin{aligned} &f(x)=\left\\{\begin{array}{ll} x^{2} & \text { if } x<0 \\ x+1 & \text { if } x \geq 0 \end{array}\right.\\\ &f(-2), f(-1), f(0), f(1), f(2) \end{aligned}$$

A function is given. (a) Find all the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. $$U(x)=x \sqrt{6-x}$$

When we seek a minimum or maximum value of a function, it is sometimes easier to work with a simpler function instead. (a) Suppose $$g(x)=\sqrt{f(x)}$$ where \(f(x) \geq 0\) for all \(x .\) Explain why the local minima and maxima of \(f\) and \(g\) occur at the same values of \(x .\) (b) Let \(g(x)\) be the distance between the point \((3,0)\) and the point \(\left(x, x^{2}\right)\) on the graph of the parabola \(y=x^{2} .\) Express \(g\) as a function of \(x\) (c) Find the minimum value of the function \(g\) that you found in part (b). Use the principle described in part (a) to simplify your work.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

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.