Step 1: Alternative form
Notice that $5 + 4\sqrt{x} + x = (\sqrt{x} + 2)^2 + 1$.
Step 2: Write in t
With $\sqrt{x} = 1 - t$, $\sqrt{x} + 2 = 3 - t$, so $f(t) = (3 - t)^2 + 1$.
Step 3: Evaluate
$f(6) = (3 - 6)^2 + 1 = 9 + 1 = 10$.
Step 4: Cross-check
Expanding $(3-t)^2 + 1 = t^2 - 6t + 10$, which matches the other route.
Step 5: Another value check
We can confirm using a specific $x$. Take $x = 4$, so $\sqrt{x} = 2$ and $g(4) = -1$. Then $f(-1) = 5 + 8 + 4 = 17$, and the polynomial $t^2 - 6t + 10$ at $t = -1$ gives $1 + 6 + 10 = 17$. This matches, so the formula for $f$ is right and $f(6) = 10$.
Final Answer:
f(6) equals 10. This is option (B).
\[ \boxed{\text{(B) }10} \]