## anonymous 4 years ago Conceptual Problem

1. anonymous

$$\mathsf{\textbf{Question : }}$$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.

2. anonymous

wtf is $$f\circ g(x)$$? is $$\circ$$ multiplication?

3. anonymous

No, it's for composite function. $f \circ g(x) = f\left( g(x)\right)$

4. anonymous

oh ok i know that one but never seen it expressed with $$\circ$$

5. anonymous

Oh, hmm this question was asked on math.stackexchange. I copied the whole LaTeX and question from there, and pasted it here.

6. mathmate

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.

7. anonymous

i am confused right away because it is not true that $f\circ g = g\circ g$

8. anonymous

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

9. anonymous

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 $g\circ f(x)=g(x^2+2x)=(x^2+2x)^3=x^6+6 x^5+12 x^4+8 x^3$

10. anonymous

we solve for x by subtracting x from both sides???? don't give this a second thought, it is gibberish

11. anonymous

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

12. anonymous

$(x^2+2x)^3=0\implies x^2+2x=0\implies x =0 \text{ or } x=-1$

13. anonymous

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'

14. anonymous

yes i see the answer, the third one from alex becker. it is crap, ignore it

15. anonymous

I can't see alex becker's answer ?!

16. anonymous

He deleted it probably.

17. anonymous

But I have posted it on the first post in this thread. You can read it.

18. anonymous

Aha, he probably realized.