Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the equation \(x^{5}+4 x^{3}-7 x+14=0\) has at least one real solution.

Step-by-Step Solution

Answer

By the Intermediate Value Theorem, there is a real solution between \(-2\) and \(2\).

Step 1: Understand the Equation

Given the equation \(x^{5}+4x^{3}-7x+14=0\), we need to show that there is at least one real solution. This means that there exists some value \(x = c\) such that the equation equals zero.

Step 2: Use the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a continuous function \(f(x)\) takes on values of opposite signs at two points \(a\) and \(b\), then there is at least one root, \(c\), between \(a\) and \(b\). "Root" here means \(f(c) = 0\).

Step 3: Check Continuity of \(f(x)\)

The function \(f(x) = x^{5}+4x^{3}-7x+14\) is a polynomial, which is continuous everywhere on the real number line.

Step 4: Evaluate \(f(x)\) at Specific Points

We will evaluate \(f(x)\) at two points. Let's try \(x = -2\) and \(x = 2\):\[f(-2) = (-2)^{5}+4(-2)^{3}-7(-2)+14 = -32 + (-32) + 14 + 14 = -36\]\[f(2) = (2)^{5}+4(2)^{3}-7(2)+14 = 32 + 32 - 14 + 14 = 64\]We notice that \(f(-2) = -36\) and \(f(2) = 64\).

Step 5: Conclude with IVT

Since \(f(-2) = -36\) is negative and \(f(2) = 64\) is positive, by the Intermediate Value Theorem, there must be at least one root between \(-2\) and \(2\). Therefore, the equation has at least one real solution.

Key Concepts

Polynomial Functions

Polynomial functions are equations that involve terms made up of variable components raised to various powers and constant coefficients. The general form of a polynomial function is given as \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0\), where the coefficients \(a_n, a_{n-1}, \ldots, a_0\) are real numbers and the exponents are non-negative integers.

Key properties of polynomial functions include:

• They are defined for all real numbers \(x\).
• They are differentiable everywhere in their domain.
• The degree of the polynomial (the highest power of \(x\)) dictates the general shape of the graph of the function.

Understanding these properties helps in analyzing and predicting the behavior of polynomials. For example, in our equation \(x^5 + 4x^3 - 7x + 14 = 0\), it is a fifth-degree polynomial. This means it can have up to 5 real roots. However, by using the Intermediate Value Theorem, we concluded the presence of at least one real solution.

Continuity

Continuity in mathematics generally refers to the property of a function to behave predictably without any jumps, breaks, or holes at any points in its domain. A function is considered continuous if, at any chosen points, the value of the function continuously approaches the limit. This can be mathematically expressed as \(f(a) = \lim_{x \to a} f(x)\) for any point \(a\) in the domain.

Polynomial functions, like the one in our exercise, are continuous everywhere on the real number line. This continuity is crucial because it allows us to use powerful mathematical tools like the Intermediate Value Theorem to determine the existence of real solutions. For a function to apply the IVT effectively, it must be continuous over a given interval. As our polynomial is continuous, this property set the stage to apply the theorem between the interval \([-2, 2]\).

Without continuity, the Intermediate Value Theorem would not be applicable, and proving the existence of real solutions would become significantly more complicated.

Real Solutions

Real solutions of a polynomial are the values of \(x\) that satisfy the equation \(f(x) = 0\). For every real solution, the graph of the polynomial intersects the \(x\)-axis at those points. The Fundamental Theorem of Algebra tells us that a polynomial of degree \(n\) has exactly \(n\) roots, though some or all of these roots can be complex.

Our goal in this exercise is to prove at least one real solution for the equation \(x^5 + 4x^3 - 7x + 14 = 0\). Using the Intermediate Value Theorem, we leveraged the evaluations at specific points, \(f(-2)\) and \(f(2)\), to show that the function changes signs, indicating the presence of a real root. Thus, we confirmed there is at least one real solution in the interval \([-2, 2]\).

• Real solutions correspond to physical, meaningful values when interpreting polynomials in applied scenarios.
• Unlike complex roots, real solutions directly relate to intersections with the real number line.

In real-world applications, identifying real solutions helps solve practical problems, whether calculating break-even points in economics or determining time or rate in physics. By proving a real solution, we provide foundational support for further analysis or practical use of the polynomial model.