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If $f,g\colon\mathbb{R}\to\mathbb{R}$ have the following properties:

i) $f(g(x))=x^2-3x+4 $

ii) $g(f(2))=2 $

Determine at least a Real solution for the equation $f(x)=g(x)$

Choose the right answer:

$a)x=1$

$b)x=-2$

$c)x=2$

$d)x=-2$

$e)x=4$

$f)x=3$

I want to know how to solve this the practical way, not by plugging the solutions in. Thank you very much for your help!

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    @ChrisEagle: Thanks. I guess if such$f$and$g$do not exist, any answer is trivially true. I was just worried that this problem might have been in the class of problems like: f(f(x)) = x, f(f(f(x)) = x + 1 In which case there are no solutions.2012-05-25

2 Answers 2

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First we can compute that $f(2)=f(g(f(2)))=f(2)^2-3\cdot f(2)+4$. Now it follows that $f(2)^2-4f(2)+4=0$, or equivalently that $(f(2)-2)^2=0$. This shows that $f(2)=2$. Since $f(2)=2$, it follows that $g(2)=g(f(2))=2$, so we conclude that $f(2)=g(2)$.

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    Thank you very much for the complete answer!2012-05-25
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We have $f(2)=f(g(f(2)))=(f(2))^2-3f(2)+4$. Thus $f(2)$ is a solution of the equation $x=x^2-3x+4$. But this simplifies to $x^2-4x+4=0$ and then $(x-2)^2=0$, so in fact $f(2)=2$. Thus $2=g(f(2))=g(2)$ and so $f(2)=g(2)=2$.

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    Nice, I was trying to figure out how to word this.2012-05-25