Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
1. Limits and Continuity
Continuity
Problem 2.5.54
Textbook Question
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.

1
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.
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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.
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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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