Is it a polynomial?

Let $P(x)$ be the statement $f(f(x)) = x^2 - x + 1$ .

First, we will find the value of $f(1)$ . Suppose $f(1) = a$ . From $P(1)$ , we have

$\begin{aligned} f(a) &= f(f(1)) \\ &= 1^2 - 1 + 1 \\ &= 1 \end{aligned}$

From $P(a)$ , we have

$\begin{aligned} f(f(a)) &= a^2 - a + 1 \\ f(1) &= a^2 - a + 1 \\ a &= a^2 - a + 1 \\ 0 &= a^2 - 2a + 1 \\ 0 &= (a-1)^2 \\ a &= 1 \end{aligned}$

Thus $f(1) = a = 1$ .

Now we move to $f(0)$ . Suppose $f(0) = b$ . Then, from $P(0)$ , we have

$\begin{aligned} f(b) &= f(f(0)) \\ &= 0^2 - 0 + 1 \\ &= 1 \end{aligned}$

From $P(b)$ , we have

$\begin{aligned} f(f(b)) &= b^2 - b + 1 \\ f(1) &= b^2 - b + 1 \\ 1 &= b^2 - b + 1 \\ 0 &= b^2 - b \\ 0 &= b(b-1) \end{aligned}$

Thus, $b = 0$ or $b = 1$ . However, if $b = 0$ , then we have $f(0) = 0$ , so $f(f(0)) = f(0) = 0$ , contradicting $P(0)$ stating that $f(f(0)) = 1$ . Thus $f(0) = b = \boxed{1}$ , which we can verify to work.

This solution is actually incomplete: it hasn't been proven that $f$ exists, only that $f(0) = 1$ for any satisfying $f$ .

Putting $x=0$ We get $f(f(0))=1$

But $f(f(0))=f(0)^2-f(0)+1=1\implies f(0)\left(f(0)-1\right) =0$

So $f(0)=1,0$ ,but $f(0)$ Can not be $0$ ,otherwise there will be divergence in the mapping of the function, so $f(0)=1$

Now we put f(x)=y, and we have that y∈R. Hence, we can write f(f(y))=y^2-y+1, but f(y)=f(f(x))=x^2-x+1 So we get f(x^2-x+1)=〖f(x)〗^2-f(x)+1 Now f(1)=〖f(1)〗^2-f(1)+1 by setting x=1 yielding f(1)=1 And set x=0 gets f(0)=0 or f(0)=1, but if f(0)=0 then f(f(0))=f(0)=1 contradiction.

$f(x) = ax^{2} + bx + c$

$f(f(x)) = a (ax^{2} + bx + c) + b (ax^{2} + bx + c) + c$

$f(f(x)) = (a^{2} + ab) x^{2} + (ab + b^{2}) x + ac + bc + c$

We know that $f(f(x))= x^{2} - x + 1$ so then we should have:

$a^2 + ab = 1 => a(a + b) = 1$

$ab + b^2 = -1 => b(a + b) = -1$

$ac + bc + c = 1 => c(a + b + 1) = 1$

From the first two we will have a = -b

Using this in the third we will have c = 1

$f(x) = ax^{2} - ax + 1$

$f(0) = \boxed{1}$

f(f(x)) = $x^2$ - x - 1 Here, if we take the inner f(x)= a,

f(a) = 1

So, a = f(x) = 1 , no matter what the value of x is, f(x) will always be 1. You could think of it as a graph of x versus y where y = f(x) = 1, hence a line parallel to the x axis as y = 1.

Here, since x = 0, f(x) = 1

(even when x= .....-5,-4,-3,-2,-1,0,1,2,3,4,5.... )

in the given equation f(f(x)) first replace x with f(x). u will get f(f(f(x)))=f(x)^2-f(x)+1. after this replace x first with 0 and then with 1 solve the quadratic and then u will get f(0) =1.

by substitution in equation f(x)=0 so f(0)=0^2-0+1=1

$I think$

Find the number a that $f(f(a)) = a$

$f(f(a)) = a^{2} - a + 1 = a$

$a^{2} - 2a + 1 = 0$

$(a-1)^{2} = 0$

$a = 1$

It means $f(f(1)) = 1$ , if we put 0 in the composit function f(f(0)) = 1 = f(1)

than f(0) = 1

Easy question for secondary (KS4/5) students.

If I rewrite the function as g(f(x))=x^2 - x + 1. I see that the function is an incomplete square. G(f(x))=x^2 -2x + 1 + x = (x - 1)^2 + x = (x - 1)^2 + (x - 1) + 1. Now I know that f(x)=x - 1. I replace it with y. Now g(y)=y^2 + y + 1. Put y = 0. You get g(y)=1.

f(f(f(0)))=f(1) but also (by assiciativity of function composition) f(f(f(0)))=f(0)^2-f(0)+1 and so f(0)^2-f(0)+1=f(1)

So it's seems worthy to find f(1) first. Employing the same technique we get:

f(1)=f(1)^2-f(1)+1 Which is solved to f(1)=1 thus: f(0)^2-f(0)+1=1 and f(0) is must be 0 or 1.

but if f(0) is 0 then f(f(0))=f(0)=0 and not 1 as it should be.

so f(0)=1

Part of me really does love the cleverness that is required to solve this problem, but I'm slightly disappointed by the fact that the answer is so easy to guess by faulty logic. Obviously, the nature of the parameters makes this aspect difficult to design around, but it still would've been nice.

The function is definitely not a polynomial, but I can't seem to prove it even exists, let alone find an explicit formulation.

first find f(0) f(0)=1 thn f(f(0))=f(1)=1

f(x) = 0-0+1 = 1 , let f(x) = y =1 , f(y) = (1^{2}) -1+1 = 1

We can see that $f$ has its inverted on $(-\infty,\frac{1}{2})$ because suppose there are $x, y\in (-\infty, \frac{1}{2})$ , $x\neq y$ and $f(x)=f(y)$ , this will lead to $0=f(f(x))-f(f(y))=(x^2-x+1)-(y^2-y+1)=(x-y)(x+y-1)\neq 0$ (contradiction)

We call $g$ the inverted of $f$ . We then have: $f(x)=g(f(f(x))=g(x^2-x+1)$ and $f(x)=f(f(g(x)))=g^2(x)-g(x)+1$

With that, we have: $f(0)=g(1)=g^2(1)-g(1)+1$

We can easily see that $f(0)=g(1)=1$ in this case

This is the right way (i think)

$f (f (0)) = f (f (1)) = 1$ by subbing into original equation

$f (0)=f (1)$

$f (x) = c, c$ is some constant

We get $f (c)= x^2 - x + 1= c$ (remember that $f(x)$ is a constant)

This implies that $x^2-x = 0 =>x = 0,x=1$ and $c=1$

$f (0) = c = 1$