/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 Consider a polynomial \(f(x)\) w... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 82

Consider a polynomial \(f(x)\) with real coefficients having the property \(f(g(x))=g(f(x))\) for every polynomial \(g(x)\) with real coefficients. Determine and prove the nature of \(f(x)\).

Short Answer

Expert verified
The nature of the function \(f(x)\) which satisfies \(f(g(x)) = g(f(x))\) for every real polynomial \(g(x)\) is \(f(x) = x\).

Step by step solution

01

Instantiate with simple cases

The first step is to test the given property with simple polynomial cases for \(g(x)\). For example, if we use \(g(x) = x\), \(f(g(x)) = g(f(x))\) simplifies to \(f(x) = f(f(x))\). This shows that any value \(x\) for which \(f(x)\) evaluates to is the same as \(f(x)\). Thus, \(f(x)\) must be of a specific form that remains unchanged when subjected to composition within itself. This suggests a linear form. Let's prove that with further steps.
02

Test for lower degree

Let's evaluate the simplest non-constant polynomial \(g(x) = x\). Evaluating, we get \(f(x) = f(f(x))\) for every \(x\). Now, suppose \(f(x)\) is a polynomial of degree \(n > 1\). Because of the fundamental theorem of Algebra, there exists \(x_1 \neq 0 \) such that \(f(x_1) = x_1\). But using \(g(x) = x + x1\), we get \(f(x + x1) = x + x1\), which contradicts the mean value theorem, because the slopes of \(f(x)\) and \(x\) are not equal for \(x \neq x1\). Hence \(f(x)\) must be linear.
03

Prove linearity

Now suppose that \(f(x) = ax + b\) where \(a, b\) are real numbers. Then testing with \(g(x) = x\) we have that \(ax + b = af(x) + b\). This gives \(a = 1, b = 0\). Thus, \(f(x) = x\) is the only solutions. Now we have to prove this is indeed a solution.
04

Validate solution

Replacing \(f(x)\) with \(x\) in the given equation, \(f(g(x))=g(f(x))\), we obtain \(g(x) = g(x)\), which is true for every polynomial \(g(x)\) with real coefficients. Hence, it is verified that if \(f(x) = x\) , then \(f(g(x)) = g(f(x))\) for any polynomial \(g(x)\), thus proving the conjecture.

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Most popular questions from this chapter

True or False? In Exercises \(50-53\), determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Use the \(\varepsilon-\delta\) definition of infinite limits to prove that \(\lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=\infty\)

In the context of finding limits, discuss what is meant by two functions that agree at all but one point.

Prove that if \(\lim _{x \rightarrow c} f(x)\) exists and \(\lim _{x \rightarrow c}[f(x)+g(x)]\) does not exist, then \(\lim _{x \rightarrow c} g(x)\) does not exist.

Use a graphing utility to graph the function on the interval \([-4,4] .\) Does the graph of the function appear continuous on this interval? Is the function continuous on [-4,4]\(?\) Write a short paragraph about the importance of examining a function analytically as well as graphically. $$ f(x)=\frac{e^{-x}+1}{e^{x}-1} $$

Sketch the graph of the function. Use a graphing utility to verify your graph. $$ f(x)=\arctan x+\frac{\pi}{2} $$
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