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Find an $f$ and a $g$ function given that $$f(g(x)) = \sqrt{1-x^2},\\ g(f(x)) = \left(\frac{x-2}{x+1}\right)^2$$

I'm a bit confused on this one. Would $g(x)$ and $f(x)$ for the two equations be represented as $x?$ And where do you go from here?

Hatmix5
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  • Note $F = f^2$ and $\phi = g(f(x))$. By substituting $x$ by $f(x)$ in the first equation and taking $f(...)$ of the second, you end up with $F + F \circ \phi = 1$. – Alexandre Halm Oct 27 '14 at 10:11
  • If you're wondering what they mean, take for instance $f(x)=\sqrt x$ and $g(x)=1-x^2$. Then we have that $f(g(x)) =\sqrt{1-x^2}$, which satisfies the first equation. However, $g(f(x)) =1-|x|$, which is not compatible with the second equation. So those are not the right $f$ and $g$. – Arthur Oct 27 '14 at 10:11
  • @Arthur So for the second equation it would be x^2 and x-2/x-1? – Hatmix5 Oct 27 '14 at 10:18
  • Yes. But you need to find an $f$ and a $g$ that works for both equations at the same time. – Arthur Oct 27 '14 at 10:21

1 Answers1

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It doesn't matter how they are represented. they can be parameterised as f(t) and g(t).

you can substitute x with f(x) in the first equation and get $$f(g(f(x))) = \sqrt{1-(f(x))^2},\\ $$

and then replace g(f(x)) in the above equation with the following:

$$g(f(x)) = \left(\frac{x-2}{x+1}\right)^2$$

You would have an equation in only one function which you can then manipulate to obtain its definition.

anu
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