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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
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.8.58d
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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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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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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Related Practice
Textbook Question

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d. Find the time when the object strikes the ground.

A payload is released at an elevation of 400 m from a hot-air balloon that is rising at a rate of 10 m/s.

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Interpreting the derivative The graph of f' on the interval [-3,2] is shown in the figure. <IMAGE>


f. Sketch one possible graph of f.

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Sketch a graph of a function f with the following properties.


f' < 0 and f" < 0, for 8 < x < 10

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{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>


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Textbook Question

Let ƒ(x) = (x - 3) (x + 3)²


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