/*! 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 112 Consider a polynomial \(f(x)\) w... [FREE SOLUTION] | 91影视

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Chapter 0: Problem 112

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 polynomial \(f(x)\) that satisfies the property \(f(g(x))=g(f(x))\) for every polynomial \(g(x)\) with real coefficients is the identity function, \(f(x) = x\).

Step by step solution

01

Substituting Identity Function

The first step is substituting the identity function, where \(g(x) = x\), into the original equation. This gives: \(f(g(x)) = g(f(x)) \Rightarrow f(x) = g(f(x)) \Rightarrow f(x) = x\), because \(g(x)\) in this case is the identity function. Therefore, the function \(f(x)\) must be linear, i.e., in the form \(f(x) = ax + b\) where \(a\) and \(b\) are some constants.
02

Substituting Constant Function

To find out the constants \(a\) and \(b\), let's use a constant function for \(g(x)\). We'll set \(g(x) = c\), where \(c\) is a constant. This leads to \(f(g(x)) = g(f(x)) \Rightarrow f(c) = g(f(c))\). Since \(g(x)\) is a constant function, \(g(f(c)) = c\). Therefore, we have \(f(c) = c\). By replacing \(c\) with an arbitrary constant, it shows that \(f(x)\) must be an identity function, so \(a = 1\) and \(b = 0\).
03

Proving the Result

To prove this result found in step 2, replace \(f(x)\) with \(x\) in the given equation \(f(g(x)) = g(f(x)) \Rightarrow x = g(x)\). This equation is clearly true for any polynomial \(g(x)\) with real coefficients. Therefore, \(f(x) = x\) is indeed the identity function that satisfies the given condition.

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Most popular questions from this chapter

Proof Prove that if the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) lie on the same line as \(\left(x_{1}^{*}, y_{1}^{*}\right)\) and \(\left(x_{2}^{*}, y_{2}^{*}\right),\) then $$ \frac{y_{2}^{*}-y_{1}^{*}}{x_{2}^{*}-x_{1}^{*}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} $$ Assume \(x_{1} \neq x_{2}\) and \(x_{1}^{*} \neq x_{2}^{*}\)

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