If $f(x)=3 f(1-x)+1$ for all $x$, the value of $f(2016)$?

Question

If $$f(x)=3 f(1-x)+1$$ for all $x$, what is the value of $f(2016)$?

I am not sure how to do this, because I see two "$f$"s.

All I could try is substituting,

$$f(x)=3(1-2016)+1\\ =-6044$$

Which I am pretty sure wrong.

How do I deal with this question? when there is $f$ around a braket?

Thank you

Answer 1

Hint: $f(2016) = 3f(-2015) + 1$. But $f(-2015)= 3f(2016)+1$. Conclude

Answer 2

We have

$$f(x)=3(3f(1-(1-x))+1)+1$$

$$=9f(x)+4$$

$\implies$

$$f(x)=-\frac12=f(2016)$$

Answer 3

f (2016)= 3f (-2015)+1

f (-2015)=3f (1-(-2015))+1=3f (2016)+1.

So f (2016)=3 (3f (2016)+1)+1=9f (2016)+4

So -8f (2016)=4

So f (2016)=-1/2.

Now the real question, is how to solve for all x.