Ch. 4 - Applications of the Derivative

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 4 - Applications of the Derivative

Problem 4.8.58a

Chapter 4, Problem 4.8.58a

{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).

a. Without using a calculator, find the values of a, with 0 ≤ a ≤ 4, such that f has a fixed point. Give the fixed point in terms of a.

Verified step by step guidance

Start by setting up the equation for a fixed point. A fixed point of the function f(x) = ax(1 - x) is a value of x such that f(x) = x. Therefore, set ax(1 - x) = x.

Rearrange the equation ax(1 - x) = x to form a quadratic equation. This can be done by expanding the left side to get ax - ax^2 = x, and then moving all terms to one side: ax - ax^2 - x = 0.

Factor the quadratic equation ax - ax^2 - x = 0. Start by factoring out x from the terms: x(a - ax - 1) = 0. This gives two potential solutions: x = 0 or a - ax - 1 = 0.

Solve the equation a - ax - 1 = 0 for x. Rearrange it to find x in terms of a: ax = a - 1, which simplifies to x = (a - 1)/a, provided a ≠ 0.

Determine the values of a for which the fixed points are valid. Since 0 ≤ a ≤ 4, check the conditions for x = 0 and x = (a - 1)/a to be valid fixed points. Note that x = 0 is always a fixed point, and x = (a - 1)/a is valid when a > 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 determine the values of a for which the function intersects the line y = x.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, typically expressed in the form f(x) = ax^2 + bx + c. In this case, the function f(x) = ax(1 - x) can be expanded to a quadratic form, which allows for the analysis of its properties, such as its vertex, axis of symmetry, and roots. Understanding the behavior of quadratics is essential for finding fixed points and analyzing their stability.

Parameter Variation

Parameter variation involves changing the values of parameters in a function to observe how the function's behavior changes. In this problem, the parameter a affects the shape and position of the quadratic function f(x). By varying a within the specified range (0 ≤ a ≤ 4), we can determine how many fixed points exist and their corresponding values, which is crucial for solving the problem.

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Textbook Question

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a. Find the time at which the object first passes the rest position, y = 0.

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a. How many people should the guide take on a tour to maximize the profit?

Textbook Question

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