Let, $x + \frac{1}{x} + 4 = t$

$\therefore x + \frac{1}{x} = t – 4$

$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {t^2} – 8t + 16$

$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} – 14 = {t^2} – 8t = t(t – 8)$

$\therefore f(t) = t(t – 8)$

So, $f(0) = 0$