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Magazine Chapter 1: Problem 38
In Exercises \(33-40,\) which of the following functions have the given property? I. \(y=\frac{x^{2}-2}{x^{2}+2}\) II. \(y=\frac{x^{2}+2}{x^{2}-2}\) III. \(y=(x-1)(1-x)(x+1)^{2}\) IV. \(y=x^{3}-x\) V. \(y=x-\frac{1}{x}\) VI. \(y=\left(x^{2}-2\right)\left(x^{2}+2\right)\) More than two distinct zeros.
Short Answer
Expert verified
Functions III and IV have more than two distinct zeros.
Step by step solution
01
Understanding the Problem
We need to identify functions from the list that have more than two distinct zeros. A zero of a function is a value of x for which the function equals zero. We will examine each function to find where it equals zero.
02
Analyzing Function I
Function I is \( y = \frac{x^2 - 2}{x^2 + 2} \). The numerator must be zero for the whole expression to be equal to zero. Solving \( x^2 - 2 = 0 \) gives \( x = \pm \sqrt{2} \). These are two zeros.
03
Analyzing Function II
Function II is \( y = \frac{x^2 + 2}{x^2 - 2} \). The numerator \( x^2 + 2 \) is always positive and cannot be zero. Hence, this function has no zeros.
04
Analyzing Function III
Function III is \( y = (x-1)(1-x)(x+1)^2 \). The zeros are found when each factor equals zero: \( x-1 = 0 \) or \( 1-x = 0 \) or \( x+1 = 0 \). This results in zeros at \( x = 1, x = -1, \) with \( x-1 \) and \( 1-x \) being equivalent at \( x = 1 \). Therefore, the zeros are \( x = -1, 1 \) (considering distinct zeros).
05
Analyzing Function IV
Function IV is \( y = x^3 - x \). This factors to \( x(x-1)(x+1) \). Setting each factor to zero gives the zeros as \( x = 0, x = 1, x = -1 \). There are three distinct zeros.
06
Analyzing Function V
Function V is \( y = x - \frac{1}{x} \). Setting this equal to zero gives \( x = \frac{1}{x} \), which solves to \( x^2 = 1 \) resulting in zeros at \( x = 1 \) and \( x = -1 \). This function has two distinct zeros.
07
Analyzing Function VI
Function VI is \( y = (x^2 - 2)(x^2 + 2) \). Solving \( x^2 - 2 = 0 \) gives roots \( x = \pm \sqrt{2} \) and \( x^2 + 2 = 0 \) has no real roots since it's always positive. This function has two distinct zeros.
08
Summary and Conclusion
Functions having more than two distinct zeros are III \((x-1)(1-x)(x+1)^2\) and IV \(x^3 - x\). Both functions III and IV have three distinct zeros, satisfying the condition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distinct Zeros
When talking about zeros of a function, we mean the values of \(x\) that make the function equal zero. If a function has multiple zeros, it may have some that are distinct and some that are repeated.
Distinct zeros are the zeros that are different from each other.
Distinct zeros are the zeros that are different from each other.
- If a polynomial has a zero at \(x = a\), then \(x\) does not equal the value that would give another root already counted; they have to be different.
- You can find distinct zeros by looking at the factors of the function. Different factors will give different zeros.
- For example, in the function \(y = (x-1)(x+2)(x+3)\), the distinct zeros are \(x = 1, -2, -3\).
Polynomial Zeros
The zeros of a polynomial function are the \(x\)-values that make the function equal zero. In other words, they are the roots or solutions to the equation formed when you set the polynomial equal to zero.
Polynomials can have several characteristics:
Polynomials can have several characteristics:
- A polynomial of degree \(n\) can have up to \(n\) zeros.
- The zeros might be real numbers or complex numbers.
- For example, the polynomial \(y = x^3 - 6x^2 + 11x - 6\) can be factored to \( (x-1)(x-2)(x-3)\), giving the zeros \(x = 1, 2, 3\).
Factorization
Factorization is a method used to solve polynomial equations, by expressing the polynomial as a product of its factors. These factors reveal the zeros of the polynomial.
Here is how factorization works:
Here is how factorization works:
- Break down a polynomial into simpler expressions called factors. For example, \(x^2 - 5x + 6\) can be factorized into \((x-2)(x-3)\).
- Each factor set to zero provides a zero of the function. So from \((x-2)(x-3)\), you get zeros \(x = 2\) and \(x = 3\).
- Multiple factors can reveal multiple zeros, and factorization is key in finding these zeros.
Real Roots
Real roots of a polynomial are the zeros that are real numbers, as opposed to complex numbers. Real roots can be thought of as points where the polynomial graph crosses or touches the x-axis. Here's what to consider:
- Real roots are easy to visualize, as you can usually see them on a graph of the polynomial.
- Not all polynomial equations have real roots; some might have complex roots or none at all.
- For example, the polynomial \(x^2 + 1 = 0\) has no real roots because no real number squared equals \(-1\). Instead, its roots are complex numbers.
