/*! 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 38 Determine whether the equation d... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 38

Determine whether the equation defines y as a function of \(x .\) \(x+y^{2}=1\)

Short Answer

Expert verified
No, y is not a function of x.

Step by step solution

01

Rewrite the Equation

The given equation is \(x + y^2 = 1\) . To determine if y is a function of x, we need to solve for y in terms of x.
02

Isolate y

Rearrange the equation to isolate the term involving y on one side: \( y^2 = 1 - x \) .
03

Solve for y

Take the square root of both sides of the equation to solve for y: \( y = \pm \sqrt{1 - x} \) . This means that for a single value of x, there are two corresponding values of y.
04

Analyze the Result

Since for a given value of x, there are two possible values for y, this violates the definition of a function. Thus y is not defined as a function of x.

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Key Concepts

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

Solving Equations
To determine if the given equation defines y as a function of x, we first need to solve the equation for y. This process is called 'solving equations'. We start by isolating the term that contains y. In our example, the equation is \( x + y^2 = 1 \). To isolate \( y \), we subtract \( x \) from both sides, leading to \( y^2 = 1 - x \). The final step involves taking the square root of both sides to get \( y \). Hence, \( y = \pm \sqrt{1 - x} \). However, this split into both positive and negative square roots shows there are multiple values of y for each x, which is crucial when we discuss the definition of a function.
Function Definition
In mathematics, a function is a relationship between two variables where each input (x) is associated with exactly one output (y). To determine if a relationship is a function, we look at the equation and its outputs. When the equation \( x + y^2 = 1 \) is solved for y, we obtain \( y = \pm \sqrt{1 - x} \). This means for every single x, there can be two different values of y, one positive and one negative. Because a specific x does not correspond to a unique y in this equation, this relationship is not considered a function. Thus, the equation does not define y as a function of x.
Square Roots
Square roots are fundamental in solving equations involving squares. The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because \( 3 \times 3 = 9 \). However, every positive number has two square roots: one positive and one negative, denoted as \( \pm \sqrt{number} \). In our equation, after isolating y and solving \( y^2 = 1 - x \), we need to take the square root of both sides. This gives us \( y = \pm \sqrt{1 - x} \). The presence of both a positive and a negative root shows that for each x, there can be two possible y values, which affects our determination of function.

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Most popular questions from this chapter

The relationship between the Celsius \(\left({ }^{\circ} \mathrm{C}\right)\) and Fahrenheit \(\left({ }^{\circ} \mathrm{F}\right)\) scales for measuring temperature is given by the equation $$ F=\frac{9}{5} C+32 $$ The relationship between the Celsius \(\left({ }^{\circ} \mathrm{C}\right)\) and \(\mathrm{Kelvin}(\mathrm{K})\) scales is \(K=C+273 .\) Graph the equation \(F=\frac{9}{5} C+32\) using degrees Fahrenheit on the \(y\) -axis and degrees Celsius on the \(x\) -axis. Use the techniques introduced in this section to obtain the graph showing the relationship between Kelvin and Fahrenheit temperatures.

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Can a function be both even and odd? Explain.

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Ethan has $$\$ 60,000$$ to invest. He puts part of the money in a CD that earns \(3 \%\) simple interest per year and the rest in a mutual fund that earns \(8 \%\) simple interest per year. How much did he invest in each if his earned interest the first year was $$\$ 3700$$.
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