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Chapter 0: Problem 45

Deciding Whether an Equation Is a Function In Exercises \(43-46,\) determine whether \(y\) is a function of \(x\). $$y^{2}=x^{2}-1$$

Short Answer

Expert verified
No, \(y\) is not a function of \(x\) in the given equation \(y^{2}=x^{2}-1\).

Step by step solution

01

Solve the Equation for Y

To solve for y in the given equation \(y^{2}=x^{2}-1\), you would first take the square root of both sides of the equation to get y. This gives two possible solutions: \(y = \sqrt{x^{2} - 1}\) and \(y = -\sqrt{x^{2} - 1}\). This occurs because when you take the square root of a square, you have to consider both the positive and negative roots.
02

Determine Whether Y is a Function of X

Next, check if every x value corresponds to only one y value. Given that for each value of x there are two possible values of y (one positive and one negative), it is clear that y is not a function of x since the definition of a function is violated here (each input x has exactly two outputs y).

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Key Concepts

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

Domain
The domain of a function represents all the possible input values (usually x-values) that allow the function to work without any errors. When determining the domain, it's important to think about what restrictions might exist.
For the equation given, if we were looking for the domain of the expression when solved for y (i.e., considering both possible solutions for y from the step-by-step solution), we would focus on the expression under the square root: \(x^2 - 1\).
  • If \(x^2 - 1\) is negative, the square root is not defined for real numbers (as it would result in an imaginary number). To avoid this, set \(x^2 - 1 \geq 0\).
  • Solving \(x^2 - 1 \geq 0\) gives \(x \leq -1\) or \(x \geq 1\). This means the domain includes all real numbers except for values between -1 and 1 (excluding -1 and 1).
Range
While the domain concerns the inputs, the range of a function concerns the possible outputs (y-values). For any function, after determining which x-values are acceptable, we observe what y-values they produce. Since our equation is not technically a function, discussing its range might be slightly different.
When considering the equation \(y^2 = x^2 - 1\), both positive and negative y-values are possible for each x-value beyond -1 and 1. The y-values themselves depend on \(\sqrt{x^2 - 1}\).
  • The smallest value for any y will be 0 when \(x = \pm1\). This is because it's the only point where \(\sqrt{x^2 - 1}\) equals zero.
  • There is no maximum y-value since as x increases or decreases past -1 or 1, the value of \(\sqrt{x^2 - 1}\) grows indefinitely. Therefore, the range of y is all real numbers \(y \geq 0\) and \(y \leq 0\) (in the case of negative y).
  • As such, the range is "all real numbers."
Definition of a Function
A function is a specific type of relation between inputs and outputs. To qualify as a function, for each input (usually denoted as x), there must be exactly one output (denoted as y). This "one-to-one" idea is what distinguishes functions from more general equations.
Let's dissect our given equation, \(y^2 = x^2 - 1\). After rearranging and solving, we discovered two potential y-values for each x-value. Specifically, both \(y = \sqrt{x^2 - 1}\) and \(y = -\sqrt{x^2 - 1}\) are derived.
  • Given an x like 2, we encounter two possible solutions for y: \(\sqrt{3}\) and \(-\sqrt{3}\).
  • This violates the function definition since there is more than one y for an x-value.
  • Therefore, the equation doesn't represent a function, despite appearing algebraically neat.
Square Root Property
The square root property is essential in finding roots of equations of the form \(y^2 = a\). This property allows us to find two solutions: the positive and the negative roots of the number or expression under the square root.
In the exercise \(y^2 = x^2 - 1\), applying the square root property gives us \(y = \sqrt{x^2 - 1}\) and \(y = -\sqrt{x^2 - 1}\).
  • This property ensures completeness in solving 鈥� we don't overlook any potential solutions.
  • Ignoring the negative solution could lead to missing parts of understanding the full behavior of the equation or function.
  • Therefore, always consider \(\pm\) results in square root equations unless constrained otherwise.

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Most popular questions from this chapter

Ferris wheel. The model for the height \(h\) of a Ferris wheel car is \(h=51+50 \sin 8 \pi t\) where \(t\) is measured in minutes. (The Ferris wheel has a radius of 50 feet.) This model yields a height of 51 feet when \(t=0\) . Alter the model so that the height of the car is 1 foot when \(t=0\) .

Testing for Symmetry In Exercises \(29-40\) , test for symmetry with respect to each axis and to the origin. \(|y|-x=3\)

Modeling Data The table shows the populations \(y\) (in millions) of the United States for 2009 through \(2014 .\) The variable \(t\) represents the time in years, with \(t=9\) corresponding to \(2009 . \quad\) (Source: \(U . S .\) Census Bureau) $$\begin{array}{|c|c|c|c|c|c|c|}\hline t & {9} & {10} & {11} & {12} & {13} & {14} \\ \hline y & {307.0} & {309.3} & {311.7} & {314.1} & {316.5} & {318.9} \\\ \hline\end{array}$$ (a) Plot the data by hand and connect adjacent points with a line segment. Use the slope of each line segment to determine the year when the population increased least rapidly. (b) Find the average rate of change of the population of the United States from 2009 through \(2014 .\) (c) Use the average rate of change of the population to predict the population of the United States in \(2025 .\)

Finding Composite Functions In Exercises \(63-66,\) find the composite functions \(f^{\circ} g\) and \(g \circ f\) . Find the domain of each composite function. Are the two composite functions equal? $$\begin{array}{l}{f(x)=\frac{3}{x}} \\ {g(x)=x^{2}-1}\end{array}$$

Proof Prove that if the slopes of two nonvertical lines are negative reciprocals of each other, then the lines are perpendicular.
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