Ch. 2 - Limits and Continuity

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 2 - Limits and Continuity

Problem 2.5.54

Chapter 2, Problem 2.5.54

A function value Show that the function F(x) = ( x − a)²(x − b)² + x takes on the value (a + b)² for some value of x.

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First, understand that we need to show that the function \( F(x) = (x - a)^2(x - b)^2 + x \) takes on the value \( (a + b)^2 \) for some value of \( x \). This means we need to find \( x \) such that \( F(x) = (a + b)^2 \).

Set the equation \( (x - a)^2(x - b)^2 + x = (a + b)^2 \) and rearrange it to form a new equation: \( (x - a)^2(x - b)^2 + x - (a + b)^2 = 0 \).

Consider this as a new function \( G(x) = (x - a)^2(x - b)^2 + x - (a + b)^2 \). We need to find the roots of \( G(x) \), i.e., values of \( x \) for which \( G(x) = 0 \).

Analyze the behavior of \( G(x) \) by considering its derivative \( G'(x) \) to find critical points, which might help in identifying where \( G(x) \) changes sign or equals zero.

Evaluate \( G(x) \) at specific points or use the Intermediate Value Theorem, which states that if \( G(x) \) is continuous on an interval \([c, d]\) and \( G(c) \) and \( G(d) \) have opposite signs, then there exists at least one \( x \) in \( (c, d) \) such that \( G(x) = 0 \). This will help confirm the existence of such an \( x \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polynomial Functions

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In this case, F(x) is a polynomial of degree four, which means it can have up to four roots. Understanding the behavior of polynomial functions, including their continuity and the Intermediate Value Theorem, is essential for analyzing the function's values.

Roots and Values of Functions

The roots of a function are the values of x for which the function equals zero. To show that F(x) takes on the value (a + b)², we need to analyze the function's behavior and find specific x-values that yield this output. This involves setting up the equation F(x) = (a + b)² and determining if there are solutions within the domain of the function.

The Intermediate Value Theorem

The Intermediate Value Theorem states that if a function is continuous on a closed interval [c, d], then it takes on every value between F(c) and F(d). This theorem is crucial for proving that F(x) achieves the value (a + b)², as it allows us to conclude that if F(c) < (a + b)² < F(d) for some c and d, then there exists at least one x in (c, d) such that F(x) = (a + b)².

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