Problem 16 - College Algebra

Question

For the following exercises, determine whether the relation represents \(y\) as a function of \(x\). $$ x=\sqrt{1-y^{2}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Equation Structure

Our given equation is \(x = \sqrt{1-y^{2}}\). To determine if this relation represents \(y\) as a function of \(x\), we must be able to express \(y\) in terms of \(x\).

Step 2: Rearrange the Equation

Square both sides of the equation to eliminate the square root: \(x^2 = 1 - y^2\).

Step 3: Solve for \(y^2\)

Reorganize the equation to solve for \(y^2\): \(y^2 = 1 - x^2\).

Step 4: Solve for \(y\)

Take the square root of both sides to solve for \(y\): \(y = \pm \sqrt{1 - x^2}\).

Step 5: Analyze the Expression

The expression \(y = \pm \sqrt{1 - x^2}\) indicates that for each \(x\), there are potentially two values for \(y\) (one positive and one negative).

Step 6: Determine if Relation is a Function

Since each \(x\) could correspond to multiple \(y\) values, the relation \(x = \sqrt{1-y^{2}}\) does not define \(y\) as a function of \(x\).

Key Concepts

Relations in Mathematics

In mathematics, understanding relations is crucial. A relation in mathematics links elements from one set, often called the domain, to another set, called the range. For functions in algebra, this relation becomes a bit more specific. A function is a special type of relation where each input from the domain is linked to exactly one output from the range. This uniqueness criterion is essential in defining functions.
When examining whether a relationship represents a function, particularly in algebraic contexts, we should always check if every input (or "x" value) corresponds to only one output (or "y" value). If even a single input maps to more than one output, the relation cannot be considered a function.
In our exercise, examining the relation given by the equation:

• Step 5 clearly showed that the expression leads to multiple possible values for output ("y") for any given input ("x")

Thus, it confirms that the relationship is indeed not a function. The study of each component in this relation helps to clearly confirm our conclusion.

Square Root Function

Square root functions are important in algebra and mathematics. They involve finding a number which, when multiplied by itself, gives the original number. Square root operations are often represented with the radical symbol \(\sqrt{}\).
Specifically, in the equation \(x = \sqrt{1 - y^2}\), the square root introduces a layer of complexity because when solving for \(y\), you'd get \(y = \pm \sqrt{1 - x^2}\). This is because squaring a number, whether positive or negative, will yield a positive result. As a result, the square root function cannot be neglected when checking for function definition due to this duality of values.
Square roots in functions can constrain the domain and the range. This is because, typically, square roots of negative numbers are not real in the simple functions considered in elementary algebra. Therefore, you need to make sure the expression under the square root is non-negative.

Solving Equations

Solving equations often requires a systematic process. Let's examine how we solve the equation \(x = \sqrt{1 - y^2}\).
Start by understanding the given equation. You identify the structure and then work to isolate one variable.

• To free the variable from the square root, we square both sides: \(x^2 = 1 - y^2\)
• Reorganize the terms to isolate \(y^2\): \(y^2 = 1 - x^2\)
• Finally, solve for \(y\) by taking the square root of both sides, considering both positive and negative square roots: \(y = \pm \sqrt{1 - x^2}\)

This equation-solving technique exemplifies how we manage and manipulate algebraic structures to uncover solutions. Remember, always be careful handling square roots, as they indicate potential multiple solutions, as seen in this exercise. This was key to understanding why the relation does not define a function.