$\begin{array}{l}f\left( {x + \frac{1}{x} + 4} \right) = {x^2} + \frac{1}{{{x^2}}} + 16\\f(17) = ?\end{array}$

Using the substitution $x + \frac{1}{x} + 4 = t$, we have

$\begin{array}{l}x + \frac{1}{x} = t - 4\\ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {(t - 4)^2}\\ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 16 = {(t - 4)^2} + 14\end{array}$

Now, $f(t) = {(t - 4)^2} + 14$

$\therefore f(17) = {(17 - 4)^2} + 14 = 169 + 14 = 183$