Function

If we look closely, the degree of the polynomial will be squared on the LHS. In the RHS however, the degree remains unchanged assuming that the polynomial isn't constant. So we can safely conclude that this is a first degree linear polynomial with no constant term (convince yourself that this is true). Let's let

$f(x)=ax$

Then we see that

$a^2=6+a$

Solving, we see that $a=-2, 3$ . But because this is a function purely in the positive reals, $a=3$ . Plugging in $x=17$ , we see that $f(17)=3(17)=\boxed{51}$ .

Consideremos a função $f$ dada por $f(x) = ax + b$ , então:

$f(f(x)) = 6x + f(x) \\ a(ax + b) + b = 6x + ax + b \\ a^2x + (ab + b) = (6 + a)x + b \\\\ \begin{cases}a^2 = 6 + a \\ ab + b = b \end{cases}$

Equação I:

$a^2 = 6 + a \\ a^2 - a - 6 = 0 \\ (a - 3)(a + 2) = 0 \\ a = - 2 \; \text{e} \; \boxed{a = 3}$

Equação II:

$ab + b = b \\ 3b = 0 \\ \boxed{b = 0}$

Portanto,

$f(x) = 3x \\ f(17) = 3 \cdot 17 \\ \boxed{\boxed{f(17) = 51}}$

I made a trial exercise with a polynomial linear in 'x' and found that the same form arose as given in the problem. Then I used the following brute force.

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from sympy import *

x = symbols ('x')

f = Function ( 'f')

a,b=symbols('a b')

def f ( x ) :

    return a*x + b

result = (f(f(x)) - f(x)).expand()

print result

f^2(x)=6x+f(x), so f^2-f-6=0, (f+2)(f-3)=0, f(x)=3x or =-2x rej by constraint

Take f(x) = 2x+x.

f(x)= ax, so f(17)=3*17=51 :)

I was just basing on the choices.., or whatever