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I have this: $$ f\colon \mathbb{R} \to \mathbb{R}, $$ $$ f(f(x)) = x^2 - x + 1 $$ I need to show that $f(1) = 1$ and I need to show that $g(x) = x^2 - xf(x) + 1$ is not an one-to-one fuction.

I know how to solve the second problem but I have no idea how to find $f(x)$. Any help is appreciated

Mohsen Shahriari
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Rrjrjtlokrthjji
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  • Your subject says "find $f(x)$, but the questions you've actually asked are really properties about $f$. Do you need to find $f$, or just prove these properties? – Thomas Andrews Oct 05 '12 at 13:01
  • I need to prove that f(1) = 1 so I have to find f(x) first :) – Rrjrjtlokrthjji Oct 05 '12 at 13:06
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    No, you don't, as the answers below show - you can find individual values of $f(x)$ without knowing how to compute all the values of $f(x)$. (For example, $f$ might not even exist, but we can prove that if $f$ exists, then $f(1)=1$.) – Thomas Andrews Oct 05 '12 at 13:08
  • @Thomas Andrews Does your response and other below ⊔ at https://math.stackexchange.com/a/1996201/946772 lead to an answer for just the weaker conditional question to which you alluded (i.e., $\underset{ℛ=[∀⟮ | □( ∈[ℕ_0^1\small⊆{\underset{⊇\underset{ℝ}{\operatorname{dom}()}}{\operatorname{preimage}}()}]⟯]}{(_ℛ)}$), or perhaps to also the stronger broad-form question posed (i.e.: $\underset{∀⟮| ◇(x∈ℝ)⟯}{\operatorname{}()}$ suject to maximally-ascertainable definable conditions unto $$ or maybe family of functions satisfying relation )? – user946772 Jan 20 '22 at 19:48
  • @MohsenShahriari one decade later... I even got a Mathematics degree now. Desides, this question is flagged as "duplicate" of a same question posted 4 years later... – Rrjrjtlokrthjji Jan 23 '22 at 21:46
  • @Rrjrjtlokrthjji I chose to mark this post as a duplicate of the other post and not the other way around only because the way the questions are presented. The other one explicitly asks for the value of $ f $ at a specific point ($ 0 $ instead of $ 1 $, but the answers manage to find $ f ( 1 ) $, too), while here finding such a value is mixed up with finding $ f ( x ) $ for all $ x $, as discussed in the above comments. See here for a discussion of complications about finding $ f ( x ) $ for all $ x $. – Mohsen Shahriari Jan 23 '22 at 22:26
  • What I intended was only to lead the interested users directly to the more "to the point" post. Sorry if I have (unintensionally) given a negative signal of any sort with the marking. These aside, while I didn't check the dates of posting, I suppose when there are already two similar answered posts on the site, marking one as a duplicate of the other one with higher "quality" of any sort can be a good practice, and checking the dates is a secondary priority compared to that. Sorry again for any trouble. – Mohsen Shahriari Jan 23 '22 at 22:26
  • @MohsenShahriari no worries! – Rrjrjtlokrthjji Jan 23 '22 at 22:53

2 Answers2

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Hint for the first part, use:

$$f(x^2-x+1) = f(f(f(x))) = f(x)^2 - f(x) + 1$$

Thomas Andrews
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For the first part, take:

$\forall x \in \mathbb{R}: f(x)^2-f(x)+1 = (f\circ f)(f(x)) = f \circ (f\circ f)(x) = f(x^2-x+1)$. Take $x=1$ and we get : $f(1)^2-f(1)+1 = f(1)$. Now factoring gives us $(f(1)-1)^2 = 0$, so $f(1) = 1$.

Stefan
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