Problem 15 - Thomas Calculus

Question

Show that the functions have exactly one zero in the given interval. \(f(x)=x^{4}+3 x+1, \quad[-2,-1]\)

Step-by-Step Solution

Answer

The function has exactly one zero in the interval \([-2, -1]\).

Step 1: Understand the Problem

We need to show that the function \(f(x) = x^4 + 3x + 1\) has exactly one zero within the interval \([-2, -1]\). This requires showing that the function changes sign over the interval and confirming it with calculus if necessary.

Step 2: Evaluate Function at Interval Endpoints

Evaluate \(f(x)\) at the endpoints \(x = -2\) and \(x = -1\). Calculate \(f(-2) = (-2)^4 + 3(-2) + 1 = 16 - 6 + 1 = 11\). Calculate \(f(-1) = (-1)^4 + 3(-1) + 1 = 1 - 3 + 1 = -1\).This shows that \(f(x)\) changes sign from positive at \(x = -2\) to negative at \(x = -1\).

Step 3: Use Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function changes sign over an interval, it must have a zero in that interval. Since \(f(x)\) is a polynomial, it is continuous. As it changes sign between \(x = -2\) and \(x = -1\), there must be at least one zero in \([-2, -1]\).

Step 4: Show No Additional Zeros

To ensure there is only one zero, check the derivative \(f'(x)\) to understand if \(f(x)\) can change behavior. Calculate \(f'(x) = 4x^3 + 3\). For \(x \in [-2,-1]\), evaluate: \(f'(-2) = 4(-2)^3 + 3 = -32 + 3 = -29\) and \(f'(-1) = 4(-1)^3 + 3 = -4 + 3 = -1\).The derivative is negative throughout the interval, so \(f(x)\) is strictly decreasing and cannot have more than one zero in \([-2, -1]\).

Step 5: Conclusion

Since \(f(x)\) changes sign between \([-2, -1]\) and is strictly decreasing with a non-positive derivative in the interval, it has exactly one zero over \([-2, -1]\).

Key Concepts

Introduction to the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus used to establish the existence of zeros within a specific interval. It applies to continuous functions, which are functions that do not "break" or "jump" anywhere within their domain. If a continuous function such as a polynomial changes its value from positive to negative over an interval, or vice versa, the IVT guarantees that there must be at least one point where the function equals zero within that interval.

In our example, we have the function \(f(x) = x^4 + 3x + 1\). When evaluated at \(x = -2\), we get a value of 11, and at \(x = -1\), we get -1. This change in sign from positive to negative as we move from \(x = -2\) to \(x = -1\) assures us, by IVT, that there is at least one zero in that interval. The polynomial \(f(x)\) is continuous because it consists only of constant powers of \(x\) and basic algebraic operations, confirming that there are no gaps or jumps.

Therefore, by applying IVT to our situation, we are assured that a zero exists somewhere between \(-2\) and \(-1\). The task then shifts to proving that there is exactly one such zero, rather than more than one.

Understanding Polynomial Functions

Polynomial functions are a cornerstone of algebra and calculus. They take the form of expressions like \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where the exponents are whole numbers and the coefficients \(a_n\) are real numbers. These functions exhibit specific properties that are crucial in mathematics, such as continuity and smoothness. These properties are key when we analyze zeros of polynomials.

In our example, \(f(x) = x^4 + 3x + 1\) is a fourth-degree polynomial. Its graph is smooth and uninterrupted, reflecting its continuous nature. This is important because the Intermediate Value Theorem relies on the function being continuous across the interval. These characteristics also allow us to use calculus tools, such as derivatives, to ascertain more about the behavior of the function across an interval.

Polynomials can have turning points – locations where the graph changes direction. These points help to determine where zeros, or roots, might lie. However, to identify these zeros precisely within a given interval, we often pair polynomial properties with calculus techniques like derivative analysis, which can provide additional insight into their behavior.

Role of Derivative Analysis in Finding Zeros

Derivative analysis is a technique used to understand the rate of change of a function. It helps in determining the behavior of functions, such as identifying intervals where a function is increasing or decreasing. This is particularly useful when scrutinizing functions for zeros, especially to ensure the presence of a single zero in a specific interval.

For the given polynomial \(f(x) = x^4 + 3x + 1\), we calculate the first derivative, \(f'(x) = 4x^3 + 3\). This derivative function reveals how \(f(x)\) changes over the interval \([-2, -1]\). By plugging in endpoints of the interval, we found that at both \(x = -2\) and \(x = -1\), \(f'(x)\) remains negative, meaning the original function is decreasing throughout this interval.

Because \(f(x)\) is strictly decreasing and only changes sign once, it confirms that there can only be one zero in the interval. If \(f(x)\) were to have more than one zero, it would require a change in direction, indicated by a change in sign of the derivative, which is non-existent here. Thus, derivative analysis not only pinpoints a zero but also ensures that no additional zeros exist within that interval.