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anonymous
 4 years ago
Conceptual Problem
anonymous
 4 years ago
Conceptual Problem

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\mathsf{\textbf{Question : }}\)Let \(f(x)=x^2+2x\) and \(g(x)=x^3\), how many different roots does \(f\circ g(x)=g\circ f(x)\) have? \(\textbf{Answer : }\) We are in a very nice situation here because \(g(x)\) is invertible, meaning \(\sqrt[3]{x}\) is uniquely defined for any real number \(x\). So we can rewrite this as \(g^{1}\circ f\circ g(x)=f(x)\), which is true if and only if \(f(x)=x\)(since both \(g\) and \(g^{1}\) are onetoone). We solve \(f(x)=x\) for \(x\) by subtracting \(x\) from both sides to get \(x^2+x=0\). This can be factored as \(x(x+1)=0\), so our two solutions are \(x=0\) and \(x=1\).\[\] \(\) I need to understand the answer, the "... which is true if and only if \(f(x)=x\)(since both \(g\) and \(g^{1}\) are onetoone)..." part.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wtf is \(f\circ g(x)\)? is \(\circ\) multiplication?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, it's for composite function. \[f \circ g(x) = f\left( g(x)\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok i know that one but never seen it expressed with \(\circ\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, hmm this question was asked on math.stackexchange. I copied the whole LaTeX and question from there, and pasted it here.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0I get the <= part, namely if f(x)=x, then g1(f(g(x)))=g1(x(g(x)))=g1(g(x))=x The => part is not so obvious to me. I only get, after all the substitutions, and raising to the third power, x^3+2 = (x+2)^3, for which the solution is x=0 or x=1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am confused right away because it is not true that \[f\circ g = g\circ g\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0matter of fact,the more i look at it the less sense it makes. first they define a specific f, namely \[f(x)=x^2+2x\] and then they seem to use f to mean something completely different, the identity function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think this might be an answer to a different question, because nothing seems to follow from what is stated. \[f(x)=x^2+2x,g(x)=x^3,f\circ g(x)=f(x^3)=x^5+2x^3\] whereas \[g\circ f(x)=g(x^2+2x)=(x^2+2x)^3=x^6+6 x^5+12 x^4+8 x^3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we solve for x by subtracting x from both sides???? don't give this a second thought, it is gibberish

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is easy enough to find the roots of both of these. \[x^5+2x^2=x^3(x^2+2)=0\] solution is x= 0 since \[x^2+2>2\] for all x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x^2+2x)^3=0\implies x^2+2x=0\implies x =0 \text{ or } x=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah umm idk but answer got upvoted 7 times. I will try get the link of the problem. http://math.stackexchange.com/questions/101586/letfxx22xandgxx3findingtherootsoffcircgxgcircfx 'Second answer'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i see the answer, the third one from alex becker. it is crap, ignore it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't see alex becker's answer ?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0He deleted it probably.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But I have posted it on the first post in this thread. You can read it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Aha, he probably realized.
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