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For \(F(x)=\frac{1}{2} x+8,\) show how iteration gives \(F(F(x))=\) \(\frac{1}{4} x+12\). Find \(F(F(F(x))]-\) also called \(F^{(3)}(x)\). The derivative of \(F^{(4)}(x)\) is ___________.

Short Answer

Expert verified
The derivative of \( F^{(4)}(x) \) is \( \frac{1}{16} \).

Step by step solution

01

Understanding the Problem

We are given the function \( F(x) = \frac{1}{2} x + 8 \) and asked to find the result of applying this function iteratively. First, we need to find \( F(F(x)) \). Then, we'll compute \( F(F(F(x))) \), also known as \( F^{(3)}(x) \), and finally find the derivative of \( F^{(4)}(x) \).
02

Find F(F(x))

To find \( F(F(x)) \), plug the function \( F(x) = \frac{1}{2} x + 8 \) into itself: \[ F(F(x)) = F\left(\frac{1}{2}x + 8\right) = \frac{1}{2}\left(\frac{1}{2}x + 8\right) + 8. \] This simplifies to \[ \frac{1}{4}x + 4 + 8 = \frac{1}{4}x + 12. \] So, \( F(F(x)) = \frac{1}{4}x + 12 \).
03

Find F^{(3)}(x)

To find \( F^{(3)}(x) \), apply \( F(x) \) to \( F(F(x)) = \frac{1}{4}x + 12 \): \[ F(F(F(x))) = F\left(\frac{1}{4}x + 12\right) = \frac{1}{2}\left(\frac{1}{4}x + 12\right) + 8. \] Simplifying gives \[ \frac{1}{8}x + 6 + 8 = \frac{1}{8}x + 14. \] Thus, \( F^{(3)}(x) = \frac{1}{8}x + 14 \).
04

Find Derivative of F^{(4)}(x)

Recognize that each iteration decreases the coefficient of \( x \) by a factor of 2 and increases the constant. For \( F^{(4)}(x) \), this continues to \( \frac{1}{16}x + C \), where \( C \) is a constant. The derivative with respect to \( x \) is the coefficient of \( x \): \( \frac{1}{16} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Composition
Function composition is a process of applying one function to the results of another. It is often expressed as \( F(G(x)) \), which means you first apply \( G \) to \( x \), and then apply \( F \) to the result of \( G(x) \). In the context of iteration, we apply the same function multiple times. For example, with the function \( F(x) = \frac{1}{2}x + 8 \), when we find \( F(F(x)) \), we compose \( F \) with itself.
In this case, we start with the function \( F(x) \) and substitute it back into itself.
  • First, compute \( F(x) = \frac{1}{2}x + 8 \).
  • Substitute \( x \) in \( F(x) \) with the entire function \( F(x) \) itself: \( F(F(x)) = F\left(\frac{1}{2}x + 8\right) \).
  • Perform the operations: \( F\left(\frac{1}{2}x + 8\right) = \frac{1}{2}\left(\frac{1}{2}x + 8\right) + 8 \), simplifying to \( \frac{1}{4}x + 12 \).
By understanding this process, you can effectively apply and simplify function compositions in mathematics.
Derivative Calculation
The derivative of a function expresses how that function changes as its input changes. It is the fundamental tool in calculus for examining rates of change. When iterating a linear function like \( F(x) = \frac{1}{2}x + 8 \), each iteration modifies the coefficient of \( x \) by reducing it by a half.
In this exercise, we find that each successive application of the function further refines the coefficient of \( x \):
  • For \( F(x) \), the coefficient is \( \frac{1}{2} \).
  • For \( F(F(x)) \), it becomes \( \frac{1}{4} \).
  • Continuing to \( F^{(3)}(x) \), it decreases to \( \frac{1}{8} \).
  • For \( F^{(4)}(x) \), it reaches \( \frac{1}{16} \).
The derivative calculation focuses mainly on the coefficient of \( x \), which represents the gradient of the function following each iteration. Thus, the derivative of \( F^{(4)}(x) \) simplifies to \( \frac{1}{16} \). This derivative signifies the diminishing rate of change in the output concerning the input after four iterations.
Linear Function Iteration
Iteration of a linear function involves repeatedly applying the function to its own output. It's akin to a loop where the output of one step serves as the input for the next. For the linear function \( F(x) = \frac{1}{2}x + 8 \), this kind of iteration causes a predictable pattern.
Each iteration demonstrates how the function systematically reduces the coefficient of \( x \). Let's outline this procedure:
  • Initially, start with \( F(x) = \frac{1}{2}x + 8 \).
  • The first iteration \( F(F(x)) \) alters it to \( \frac{1}{4}x + 12 \).
  • Proceeding to \( F^{(3)}(x) \), it simplifies to \( \frac{1}{8}x + 14 \).
  • Each time, the consistent pattern involves halving the \( x \) term and adjusting the constant term.
This pattern highlights a stable reduction method for linear functions under iteration, exhibiting both the ease and complexity such functions can display. Iteration offers a repetitive, methodical approach to problem-solving and reveals large-scale trends within the functions themselves.

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