$f(x)$ and $g(x)$ are linear functions such that for all $x$, $f(g(x)) = g(f(x)) = x$. If $f(0) = 4$ and $g(5) = 17$, compute $f(2006)$.
1. Given that $f(x)$ and $g(x)$ are linear functions such that $f(g(x)) = g(f(x)) = x$ for all $x$, it implies that $f(x)$ and $g(x)$ are inverses of each other. Therefore, we can write:
   \[
   f(g(x)) = x \quad \text{and} \quad g(f(x)) = x
   \]

2. Let $f(x) = ax + b$. Since $f(x)$ and $g(x)$ are inverses, we can express $g(x)$ in terms of $f(x)$. If $f(x) = ax + b$, then $g(x)$ must be of the form:
   \[
   g(x) = \frac{x - b}{a}
   \]

3. We are given that $f(0) = 4$. Substituting $x = 0$ into $f(x)$, we get:
   \[
   f(0) = a \cdot 0 + b = 4 \implies b = 4
   \]
   Thus, the function $f(x)$ can be written as:
   \[
   f(x) = ax + 4
   \]

4. We are also given that $g(5) = 17$. Substituting $x = 5$ into $g(x)$, we get:
   \[
   g(5) = \frac{5 - 4}{a} = 17 \implies \frac{1}{a} = 17 \implies a = \frac{1}{17}
   \]
   Therefore, the function $f(x)$ is:
   \[
   f(x) = \frac{1}{17}x + 4
   \]

5. To find $f(2006)$, we substitute $x = 2006$ into $f(x)$:
   \[
   f(2006) = \frac{1}{17} \cdot 2006 + 4
   \]

6. Simplifying the expression:
   \[
   f(2006) = \frac{2006}{17} + 4
   \]

7. Calculating $\frac{2006}{17}$:
   \[
   \frac{2006}{17} = 118
   \]

8. Adding 4 to the result:
   \[
   f(2006) = 118 + 4 = 122
   \]

The final answer is $\boxed{122}$