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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.8.58d

{Use of Tech} Fixed points of quadratics and quartics Let f(x) = ax(1 -x), where a is a real number and 0 ≤ a ≤ 1. Recall that the fixed point of a function is a value of x such that f(x) = x (Exercises 48–51). 


d. Find the number and location of the fixed points of g for a = 2, 3, and 4 on the interval 0 ≤ x ≤ 1. 

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1
Understand the concept of a fixed point: A fixed point of a function f(x) is a value x such that f(x) = x. This means that when you substitute x into the function, the output is the same as the input.
Set up the equation for fixed points: For the function f(x) = ax(1 - x), we need to solve the equation ax(1 - x) = x to find the fixed points.
Rearrange the equation: Start by expanding the left side to get ax - ax^2 = x. Then, move all terms to one side to form a quadratic equation: ax - ax^2 - x = 0.
Factor the quadratic equation: Factor the equation ax - ax^2 - x = 0 to find the values of x. This can be done by factoring out x, giving x(a - ax - 1) = 0.
Solve for x: The factored equation x(a - ax - 1) = 0 gives two potential solutions: x = 0 and a - ax - 1 = 0. Solve the second equation for x to find the other fixed points, and analyze these solutions for different values of a (2, 3, and 4) within the interval 0 ≤ x ≤ 1.

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Key Concepts

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

Fixed Points

A fixed point of a function f(x) is a value x such that f(x) = x. This means that when the function is applied to this value, it returns the same value. Finding fixed points often involves solving the equation f(x) - x = 0. In the context of the given function, identifying fixed points helps understand the behavior of the function within a specified interval.
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Critical Points

Quadratic and Quartic Functions

Quadratic functions are polynomial functions of degree two, typically expressed in the form f(x) = ax^2 + bx + c. Quartic functions are of degree four, represented as f(x) = ax^4 + bx^3 + cx^2 + dx + e. The behavior of these functions, including their fixed points, can be analyzed using their graphs, derivatives, and algebraic properties, which are essential for solving the problem at hand.
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Introduction to Polynomial Functions

Interval Analysis

Interval analysis involves examining the behavior of functions within a specific range, in this case, 0 ≤ x ≤ 1. This is crucial for determining the existence and location of fixed points, as the function's behavior may vary significantly outside this interval. By restricting the analysis to a defined interval, one can apply techniques such as the Intermediate Value Theorem to ascertain the number of fixed points.
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if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.


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Textbook Question

{Use of Tech} Fixed points of quadratics and quartics Let f(x) = ax(1 -x), where a is a real number and 0 ≤ a ≤ 1. Recall that the fixed point of a function is a value of x such that f(x) = x (Exercises 48–51). 


c. Graph g for a = 2, 3, and 4.

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