Problem 116
Question
A graph of \(y=f(x)\) is shown, where \(f(x)=2 x^{5}-3 x^{4}+x^{3}-8 x^{2}+5 x+3\) and \(f(-x)=-2 x^{5}-3 x^{4}-x^{3}-8 x^{2}-5 x+3\). (a) How many negative real zeros does \(f\) have? Explain. (b) How many positive real zeros are possible for \(f ?\) Explain. What does this tell you about the eventual right-hand behavior of the graph? (c) Is \(x=-\frac{1}{3}\) a possible rational zero of \(f ?\) Explain. (d) Explain how to check whether \(\left(x-\frac{3}{2}\right)\) is a factor of \(f\) and whether \(x=\frac{3}{2}\) is an upper bound for the real zeros of \(f\).
Step-by-Step Solution
Verified Answer
The function \(f\) has 5, 3 or 1 negative real zeros. \(f\) also has 4, 2, or 0 positive real zeros. It's not possible for \(x=-1/3\) to be a rational zero of \(f\). To determine if \((x-\frac{3}{2})\) is a factor of \(f\) and if \(x=\frac{3}{2}\) is an upper bound for the real zeros of \(f\), one should substitute these values into \(f(x)\).
1Step 1: Determining number of negative real zeros
The function \(f(-x)\) is a reflection of \(f(x)\) about the y-axis. Comparing \(f(x)\) and \(f(-x)\), it is clear that any positive real zero of \(f(x)\) will become a negative real zero of \(f(-x)\). The function has 5 sign changes from \(f(x)\) to \(f(-x)\), thus it has 5 or 3 or 1 negative real zeros. Therefore, \(f(x)\) has 5, 3, or 1 negative real zeros.
2Step 2: Determining number of positive real zeros
The number of positive real zeros is determined by the number of sign changes in \(f(x)\). The function \(f(x)\) shows 4 sign changes, which means there can be 4 or 2 or 0 positive real zeros. Therefore, \(f(x)\) has 4, 2, or 0 positive real zeros. Given the degree of the polynomial is 5, the total number of real zeros is an odd number, so the combination could be 4 positive and 1 negative roots or 2 positive and 3 negative roots or 0 positive and 5 negative roots. This suggests that the right-hand behavior of the graph should be eventually going up and touch the positive part of y-axis.
3Step 3: Checking for rational zeros
The Rational Root Theorem states that any rational root, written in lowest terms, has a numerator that is a factor of the constant term, and a denominator that is a factor of the leading coefficient. Here, the constant term is 3 and the leading coefficient is 2. Thus, the possible rational roots are \(\pm\)1, \(\pm\)3, \(\pm\)1/2, and \(\pm\)3/2. \(x=-1/3\) is not among the possible rational roots.
4Step 4: Checking for a factor and upper bound
One can check whether \((x-\frac{3}{2})\) is a factor of \(f\) by setting \(x=\frac{3}{2}\) and then substituting it in \(f(x)\). If \(f(\frac{3}{2}) = 0\), then \((x-\frac{3}{2})\) is a factor of \(f\). To check whether \(x=\frac{3}{2}\) is an upper bound for the real zeros of \(f\), we can use the Upper Bound Theorem: If \(a > 0\) and \(f(a) > 0\), then \(a\) is an upper bound for the real zeros of \(f\). Thus, calculate \(f(\frac{3}{2})\) and ascertain if it is a positive value.
Key Concepts
Real ZerosRational Root TheoremSign ChangesUpper Bound Theorem
Real Zeros
Real zeros, also known as real roots, are the x-values where a polynomial function equals zero. These are the points where the graph of the function intersects the x-axis. To determine the number of negative or positive real zeros, we can use the function's behavior and other related theorems.
For the function given, the number of negative real zeros is linked with the number of sign changes when comparing the function to its reflected form. With 5 sign changes from \( f(x) \) to \( f(-x) \), we know it can have 5, 3, or 1 negative real zeros. Similarly, the number of positive real zeros is determined by the sign changes in \( f(x) \) itself, which results in possibilities of 4, 2, or 0 positive real zeros.
These possibilities arise from Descartes' Rule of Signs which suggests variations in potential zeros by decreasing in increments of 2.
For the function given, the number of negative real zeros is linked with the number of sign changes when comparing the function to its reflected form. With 5 sign changes from \( f(x) \) to \( f(-x) \), we know it can have 5, 3, or 1 negative real zeros. Similarly, the number of positive real zeros is determined by the sign changes in \( f(x) \) itself, which results in possibilities of 4, 2, or 0 positive real zeros.
These possibilities arise from Descartes' Rule of Signs which suggests variations in potential zeros by decreasing in increments of 2.
Rational Root Theorem
The Rational Root Theorem is a useful tool to find possible rational zeros of a polynomial function. This theorem states that if \( p/q \) is a root of a polynomial, then \( p \) must be a factor of the constant term, and \( q \) must be a factor of the leading coefficient.
For the polynomial \( f(x) = 2x^5 - 3x^4 + x^3 - 8x^2 + 5x + 3 \):
For the polynomial \( f(x) = 2x^5 - 3x^4 + x^3 - 8x^2 + 5x + 3 \):
- Constant term is 3: Factors are \( \pm 1, \pm 3 \).
- Leading coefficient is 2: Factors are \( \pm 1, \pm 2 \).
Sign Changes
Sign changes in a polynomial provide insight into the number of possible real zeros. By observing changes in the signs of the polynomial’s coefficients, we can determine variations in potential real zeros.
From the given polynomial, examining the coefficients:
From the given polynomial, examining the coefficients:
- The function \( f(x) \) has 4 sign changes.
- The reflected function \( f(-x) \) also has 5 sign changes.
- \( f(x) \)'s 4 sign changes suggest up to 4 positive real zeros.
- \( f(-x) \)'s 5 subtracts in steps of two, indicating 5, 3, or 1 negative real zeros.
Upper Bound Theorem
The Upper Bound Theorem is instrumental in narrowing down potential real zeros of the polynomial. It states that if a number \( a \), larger than zero, leads \( f(a) \) to be greater than zero, then \( a \) serves as an upper bound for polynomial real zeros.
In this context, to check whether \( x = \frac{3}{2} \) is an upper bound, you need to compute \( f(\frac{3}{2}) \):
In this context, to check whether \( x = \frac{3}{2} \) is an upper bound, you need to compute \( f(\frac{3}{2}) \):
- Substitute \( x = \frac{3}{2} \) into the polynomial.
- If the result is a positive number, then \( \frac{3}{2} \) is indeed an upper bound.
Other exercises in this chapter
Problem 115
Complete each polynomial division. Write a brief description of the pattern that you obtain, and use your result to find a formula for the polynomial division \
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Determine whether the statement is true or false. Justify your answer. It is possible for a fifth-degree polynomial to have no real zeros.
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Determine whether the statement is true or false. Justify your answer. It is possible for a polynomial with an even degree to have a range of \((-\infty, \infty
View solution Problem 117
Use any convenient method to solve the quadratic equation. $$4 x^{2}-17=0$$
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