Conceptual Problem

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Conceptual Problem

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\(\mathsf{\textbf{Question : }}\)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 one-to-one). 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 one-to-one)..." part.
wtf is \(f\circ g(x)\)? is \(\circ\) multiplication?
No, it's for composite function. \[f \circ g(x) = f\left( g(x)\right)\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh ok i know that one but never seen it expressed with \(\circ\)
Oh, hmm this question was asked on math.stackexchange. I copied the whole LaTeX and question from there, and pasted it here.
I get the <= part, namely if f(x)=x, then g-1(f(g(x)))=g-1(x(g(x)))=g-1(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.
i am confused right away because it is not true that \[f\circ g = g\circ g\]
matter 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
i 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 \[g\circ f(x)=g(x^2+2x)=(x^2+2x)^3=x^6+6 x^5+12 x^4+8 x^3\]
we solve for x by subtracting x from both sides???? don't give this a second thought, it is gibberish
it 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
\[(x^2+2x)^3=0\implies x^2+2x=0\implies x =0 \text{ or } x=-1\]
yeah umm idk but answer got upvoted 7 times. I will try get the link of the problem. http://math.stackexchange.com/questions/101586/let-fx-x22x-and-gx-x3-finding-the-roots-of-f-circ-gx-g-circ-fx 'Second answer'
yes i see the answer, the third one from alex becker. it is crap, ignore it
I can't see alex becker's answer ?!
He deleted it probably.
But I have posted it on the first post in this thread. You can read it.
Aha, he probably realized.

Not the answer you are looking for?

Search for more explanations.

Ask your own question