Chapter 3: Problem 9
Use the Intermediate value theorem to show that \(f\) has a zero between \(a\) and \(b\) $$f(x)=x^{5}+x^{3}+x^{2}+x+1 ; \quad a=-\frac{1}{2}, \quad b=-1$$
Short Answer
Expert verified
There is a zero of \( f(x) \) between \(-1\) and \(-\frac{1}{2}\).
Step by step solution
01
Understanding the Intermediate Value Theorem
The Intermediate Value Theorem states that if a function \( f \) is continuous on the interval \([a, b]\), and \( N \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f(c) = N \). Here, we will show that a zero of \( f \) exists between \( a \) and \( b \).
02
Checking Continuity of the Function
The function \( f(x) = x^{5} + x^{3} + x^{2} + x + 1 \) is a polynomial, and polynomials are continuous everywhere. Therefore, \( f \) is continuous on \([-1, -\frac{1}{2}]\).
03
Calculating \( f(a) \)
Substitute \( a = -\frac{1}{2} \) into \( f(x) \):\[f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^{5} + \left(-\frac{1}{2}\right)^{3} + \left(-\frac{1}{2}\right)^{2} + \left(-\frac{1}{2}\right) + 1\]Calculate each term:- \((-\frac{1}{2})^{5} = -\frac{1}{32}\)- \((-\frac{1}{2})^{3} = -\frac{1}{8}\)- \((-\frac{1}{2})^{2} = \frac{1}{4}\)- \((-\frac{1}{2}) = -\frac{1}{2}\)- Combine: \\[ f\left(-\frac{1}{2}\right) = -\frac{1}{32} - \frac{1}{8} + \frac{1}{4} - \frac{1}{2} + 1 = \frac{11}{32} > 0 \]
04
Calculating \( f(b) \)
Substitute \( b = -1 \) into \( f(x) \):\[f(-1) = (-1)^5 + (-1)^3 + (-1)^2 + (-1) + 1\]Calculate each term:- \((-1)^5 = -1\)- \((-1)^3 = -1\)- \((-1)^2 = 1\)- \((-1) = -1\)- Combine: \\[ f(-1) = -1 - 1 + 1 - 1 + 1 = -1 < 0 \]
05
Applying the Intermediate Value Theorem
Since \( f \) is continuous and \( f(-1) < 0 < f(-\frac{1}{2}) \), there must be a point \( c \) in \((-1, -\frac{1}{2})\) such that \( f(c) = 0 \). This confirms that \( f \) has a zero between \( a = -1 \) and \( b = -\frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Polynomial Functions
Polynomial functions are equations consisting of terms, each being a product of a constant and variables raised to whole number powers. The standard form of a polynomial function in one variable, say \( x \), looks like this: \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \), where \( a_n, a_{n-1}, \ldots, a_0 \) are constants, and \( n \) is a non-negative integer.
Examples of polynomial functions include linear functions (degree 1), quadratic functions (degree 2), cubic functions (degree 3), and so on.
Some key features of polynomial functions include:
Examples of polynomial functions include linear functions (degree 1), quadratic functions (degree 2), cubic functions (degree 3), and so on.
Some key features of polynomial functions include:
- They are smooth and continuous, with no gaps, jumps, or sharp points.
- The degree of the polynomial often determines the number of solutions or roots it might have.
- The coefficients and the degree influence the shape and position of the polynomial graph.
The Concept of Continuity
Continuity is a property of a function that suggests the graph of the function looks like an unbroken curve or line. For a function to be continuous at a point \( c \), three conditions must be met:
This property of being continuous makes polynomial functions helpful when using the Intermediate Value Theorem, which requires continuous functions.
- \( f(c) \) must be defined (the function has a value at \( c \)).
- The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
- The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).
This property of being continuous makes polynomial functions helpful when using the Intermediate Value Theorem, which requires continuous functions.
Exploring the Zero of a Function
The zero of a function, often called a root or solution, is a specific value \( x = c \) where the function equals zero, i.e., \( f(c) = 0 \). Finding a zero of a function is like identifying the point where its graph crosses or touches the x-axis.
For polynomial functions, finding zeros is crucial because:
For polynomial functions, finding zeros is crucial because:
- Zeros can indicate where the function changes sign (from positive to negative or vice versa).
- They help in factoring the polynomial if it's possible to express it as a product of simpler polynomials.
- Zeros can be used to solve equations involving polynomials by setting the entire polynomial expression equal to zero.