Problem 84 - College Algebra with Modeling and Visualization

Question

Determine if \(y\) is a function of \(x\). $$ y^{2}=x+1 $$

Step-by-Step Solution

Answer

No, \( y \) is not a function of \( x \) because it gives two values for each \( x \).

Step 1: Understand the Definition of a Function

A function is a relation where each input (or value of the independent variable) corresponds to exactly one output (or value of the dependent variable). For example, for each value of \( x \), there should be exactly one value of \( y \).

Step 2: Analyze the Given Equation

The given equation is \( y^2 = x + 1 \). We need to determine whether for each value of \( x \), there is exactly one corresponding value of \( y \).

Step 3: Solve the Equation for y

Solving \( y^2 = x + 1 \) for \( y \), we get \( y = \pm \sqrt{x + 1} \). This means \( y = \sqrt{x + 1} \) or \( y = -\sqrt{x + 1} \).

Step 4: Examine the Output for a Single Input

For any single value of \( x \), the equation \( y = \pm \sqrt{x + 1} \) will provide two possible values for \( y \). For example, if \( x = 3 \), then \( y = \sqrt{4} = 2 \) or \( y = -2 \).

Step 5: Conclude if y is a Function of x

Since any given value of \( x \) provides two possible values for \( y \), \( y \) is not a function of \( x \) according to the definition of a function.

Key Concepts

Relation

In mathematics, a relation is a connection or association between elements of two sets. When we talk about a relation in terms of a function, we are referring to a kind of pairing of elements where each input from one set (the domain) is related to one or more outputs in another set (the range). For example, think of a relation as a rule that takes a starting number (the input) and can potentially lead to several different results (the outputs).
Understanding the relation between variables is essential when determining if a given mathematical expression is a function. It allows us to see whether each input leads to a unique output or multiple outputs. In our example, the equation \( y^2 = x + 1 \) represents a relation because it associates each value of \( x \) with potentially two values of \( y \) – both the positive and negative square roots.
Therefore, this equation fails the "vertical line test" (a graphical way to test if an equation defines a function) because a vertical line would intersect the graph of \( x + 1 \) at more than one point on the \( y \)-axis.

Domain

The domain of a function or relation is the set of all possible starting values (inputs) for which the function or relation is defined. It is vital to determine the domain to understand what values will make sense for the equation or function we are working with. To find the domain of a relation like \( y^2 = x + 1 \), we assess the values of \( x \) that allow for real outputs of \( y \).
Since \( y^2 = x + 1 \), we need \( x + 1 \) to be non-negative to take the square root, meaning \( x + 1 \geq 0 \) or simply \( x \geq -1 \). Thus, the domain of the relation is all real numbers \( x \) such that \( x \) is greater than or equal to -1. This constraint ensures that for every value of \( x \) in the domain, the relationship is defined and can be evaluated.
In summary, the domain ensures that we only consider inputs that produce valid outputs within the scope of the given relation.

Range

The range of a relation or function is the set of all possible outputs (results) that the function or relation can produce. When assessing the equation \( y^2 = x + 1 \), determining the range involves identifying all possible values of \( y \) once the domain has been established.
In this equation, for each value of \( x \) in the domain \( x \geq -1 \), the outputs \( y \) are either \( \y = \sqrt{x+1} \) or \( \y = -\sqrt{x+1} \). This means \( y \) can take on any value both above and below zero, depending on whether you consider the positive or negative square root. Consequently, the range of the relation \( y^2 = x + 1 \) includes all real numbers. It reflects how \( y \) can vary greatly, representing both positive and negative outcomes, reinforcing the idea that this is not a function.
Understanding the range helps to define the output limits of the relation and confirms why this relation does not satisfy the requirements to be a function.