Beginner Functional Equation

Simply we can solve in the following way: f(6)=f(2+4)=f(2)f(4) =f(2)f(2+2) =f(2)f(2)f(2) =3.3.3=27 Hence f(6) = 27

First note that $3 = f(2) = f(1 + 1) = f(1)*f(1) = f(1)^{2} \Longrightarrow f(1) = \pm \sqrt{3}.$

Then $f(3) = f(2 + 1) = f(2)*f(1) = \pm 3\sqrt{3},$ and so $f(6) = f(3 + 3) = f(3)^{2} = (\pm3\sqrt{3})^{2} = \boxed{27}.$

Alternatively, we have that

$f(4) = f(2 + 2) = f(2)*f(2) = 3*3 = 9 \Longrightarrow f(6) = f(4 + 2) = f(4)*f(2) = 9*3 = \boxed{27}.$

You can also derive a general term that, for an even number $x$ , the function $f(x)$ may be written in the form of $f(x) = (\sqrt{3})^x$

f(a+b+c)=f(a)f(b)f(c)

f(6)=f(2+2+2)=f(2)f(2)f(2)=3^3=27

If f(2) = 3, then f(2+2) = 3 * 3 = 9. Now, f(6) = f( 2 + 2 + 2 ) = f(2) * f(2 + 2) = 3 * 9 = 27.

One another solution:

f(6) = f(3+3) =2 f(3);;
f(3)=f(2+1)=f(2)f(1)=3
f(1);;
f(2)=f(1+1)=2*f(1)=3;;
f(1)=3/2;;

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f(6)=2 (3 (3/2))=9