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Editorial
$f(69)=f(2 \times 34 +1)$
$f(34)=f(2 \times 17$
Again, $f(0)= 4f(0)-0$
$f: \mathbb{R} \to \mathbb{R} $
$f(2x)=4f(x)-4x$
$f(2x+1)=4f(x)+3$
$f(69)=?$
$f(69)=f(2 \times 34 +1)$
$f(34)=f(2 \times 17$
Again, $f(0)= 4f(0)-0$