/*! 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 45 Determine whether \(y\) is a fun... [FREE SOLUTION] | 91影视

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Chapter 0: Problem 45

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 since for each value of \(x\), there are two possible values for \(y\), positive and negative.

Step by step solution

01

Solving the equation

Start by solving the equation for \(y\). We have \(y^{2}=x^{2}-1\). Then, we take the square root of both sides to solve for \(y\). Remember that the square root has both a positive and a negative solution, so we get \(y=\pm\sqrt{x^{2}-1}\).
02

Checking for the function definition

Review the definition of a function: each value of \(x\) must map to exactly one value of \(y\). This is contradicted by the equation in step 1, where it is observed that for each \(x\), there are two possible values for \(y\), positive and negative. This means each value of \(x\) does not map to exactly one value of \(y\) as required by the definition of a function.

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Key Concepts

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

Solving Equations
Solving equations is a fundamental aspect of algebra. It involves finding the value of the variables that make an equation true. In the equation \( y^2 = x^2 - 1 \), our goal is to solve for \( y \). This requires isolating \( y \) on one side of the equation.

We do this by recognizing that \( y^2 \) suggests a need to take a square root to find \( y \). Taking the square root introduces two potential solutions 鈥� one positive and one negative. Thus, we find:
  • \( y = \sqrt{x^2 - 1} \)
  • \( y = -\sqrt{x^2 - 1} \)
The key takeaway here is understanding that solving for a variable like \( y \) often involves exploring both positive and negative roots when dealing with squares.
Function Definition
A function in mathematics is a relationship where each input (usually \( x \)) has one and only one output (usually \( y \)). When determining if \( y \) is a function of \( x \) in the context of \( y^2 = x^2 - 1 \), we apply the definition of a function.

For \( y \) to be a function of \( x \), every single \( x \) value must correspond to exactly one \( y \) value. However, from our equation solution, we see:
  • Each \( x \) corresponds to two \( y \) values: \( \sqrt{x^2 - 1} \) and \( -\sqrt{x^2 - 1} \).
This violates the one-to-one requirement of a function, resulting in the determination that \( y \) is not a function of \( x \) because it does not satisfy this core definition.
Square Root Property
The square root property is a crucial concept when solving equations involving squared terms. It states that if \( a^2 = b \), then \( a \) can be \( \pm \sqrt{b} \). This principle helps solve equations where a variable is squared, such as \( y^2 = x^2 - 1 \).

Applying the square root property here gives us both possible roots of \( y \):
  • Positive square root: \( y = \sqrt{x^2 - 1} \)
  • Negative square root: \( y = -\sqrt{x^2 - 1} \)
Understanding this property is vital because it acknowledges that squaring removes information about whether an initial number was positive or negative. This acknowledgment ensures that when solving such equations, both possibilities are considered, leading to a complete solution.

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Most popular questions from this chapter

Prove that the figure formed by connecting consecutive midpoints of the sides of any quadrilateral is a parallelogram.

Find the distance between the point and line, or between the lines, using the formula for the distance between the point \(\left(x_{1}, y_{1}\right)\) and the line \(A x+B y+\) \(C=0 .\) Point: \((6,2)\) Line: \(x=-1\)

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. \(f(x)=3 x-1\) \(\frac{f(x)-f(1)}{x-1}\)

The lines represented by \(a x+b y=c_{1}\) and \(b x-a y=c_{2}\) are perpendicular. Assume \(a \neq 0\) and \(b \neq 0 .\)

Determine whether the function is even, odd, or neither. Use a graphing utility to verify your result. $$f(x)=\sin ^{2} x$$
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