Problem 18
Question
For the following exercises, determine whether the relation represents \(y\) as a function of \(x\). $$ x^{2}+y^{2}=9 $$
Step-by-Step Solution
Verified Answer
No, the relation is not a function.
1Step 1: Understand the Definition of a Function
A relation is a function if for every input value (usually represented by \(x\)), there is exactly one output value (usually represented by \(y\)). This means, a function can't map a single \(x\)-value to multiple \(y\)-values.
2Step 2: Analyze the Given Equation
We are given the equation \(x^2 + y^2 = 9\). This equation resembles the equation of a circle with a radius of 3. A circle can have multiple \(y\)-values (two in most cases) for a given \(x\)-value unless it is a point exactly at the top or bottom of the circle.
3Step 3: Check for Multiple Outputs
Rearrange the equation to express \(y\) in terms of \(x\):\[y^2 = 9 - x^2\]Taking the square root of both sides gives:\[y = \pm \sqrt{9 - x^2}\]This shows for each \(x\) (within the domain \(-3 \le x \le 3\)), there are two possible values of \(y\), except when \(x = 3\) or \(x = -3\), which corresponds to the top and bottom points of the circle, respectively.
4Step 4: Conclude based on the Analysis
Since there are multiple possible \(y\) values for most \(x\) values within the domain (except the maximum and minimum points), the relation given by \(x^2 + y^2 = 9\) does not define \(y\) as a function of \(x\).
Key Concepts
Circle EquationRelation AnalysisFunction DefinitionMultiple Outputs
Circle Equation
In mathematics, a circle equation typically looks like \(x^2 + y^2 = r^2\), where \(r\) represents the radius of the circle. For the equation \(x^2 + y^2 = 9\), the radius \(r\) is 3, since \(3^2 = 9\). This equation represents a circle centered at the origin (0, 0) on the coordinate plane.
- An equation of a circle in this form always describes all the points that are at a distance of \(r\) from the center.
- A circle equation is a perfect example of a geometric layout, where every point on the curve is equidistant from the center.
Relation Analysis
Analyzing a relation involves checking how the values of \(x\) relate to values of \(y\). For equations like \(x^2 + y^2 = 9\), when you input a value for \(x\), there can be more than one possible value for \(y\).
- To better analyze, rearrange the equation: \(y^2 = 9 - x^2\).
- Taking the square root means \(y\) can equal \(\pm \sqrt{9 - x^2}\). This means two potential outputs for \(y\) for each \(x\), indicating multiple horizontal slices of the circle.
Function Definition
The definition of a function is straightforward: a function is a relation where each input corresponds to exactly one output. It’s like a rule where no two different outputs can arise from one singular input value.
- In mathematical notation, if \(f(x) = y\), for each \(x\) there should only be one \(y\).
- For example, the circle’s equation \(x^2 + y^2 = 9\) does not satisfy this condition since an input \(x\) might return two different \(y\) values.
Multiple Outputs
In the context of the circle equation, multiple outputs refer to the fact that for a given \(x\), there can be two corresponding \(y\) values.
- Taking \(x^2 + y^2 = 9\) and solving for \(y\), yields \(y = \pm \sqrt{9 - x^2}\).
- This means for each \(x\) (except at boundaries where \(x = \pm 3\), you’ll find two possible \(y\) values, understood geometrically as the top and bottom parts of the circle.
Other exercises in this chapter
Problem 18
For the following exercises, use each set of functions to find \(f(g(h(x)))\) ). Simplify your answers. $$f(x)=x^{4}+6, g(x)=x-6, \text { and } h(x)=\sqrt{x}$$
View solution Problem 18
Use each set of functions to find \(f(g(h(x)))\). Simplify your answers. \(f(x)=x^{4}+6, g(x)=x-6,\) and \(h(x)=\sqrt{x}\)
View solution Problem 18
For the following exercises, find the domain of each function using interval notation. $$ f(x)=\frac{1}{x^{2}-x-6} $$
View solution Problem 19
For the following exercises, use a graphing utility to determine whether each function is one-to-one. $$ f(x)=\sqrt{x} $$
View solution