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Answer :
We are given the recursive relation
$$
f(n+1) = \frac{1}{3} f(n)
$$
and the value
$$
f(3) = 9.
$$
To find $f(1)$, we work backwards:
1. First, express $f(3)$ in terms of $f(2)$:
$$
f(3) = \frac{1}{3} f(2).
$$
Solving for $f(2)$, multiply both sides by 3:
$$
f(2) = 3 \times f(3) = 3 \times 9 = 27.
$$
2. Next, express $f(2)$ in terms of $f(1)$:
$$
f(2) = \frac{1}{3} f(1).
$$
Again, solving for $f(1)$, multiply both sides by 3:
$$
f(1) = 3 \times f(2) = 3 \times 27 = 81.
$$
Thus, the value of $f(1)$ is $\boxed{81}$.
$$
f(n+1) = \frac{1}{3} f(n)
$$
and the value
$$
f(3) = 9.
$$
To find $f(1)$, we work backwards:
1. First, express $f(3)$ in terms of $f(2)$:
$$
f(3) = \frac{1}{3} f(2).
$$
Solving for $f(2)$, multiply both sides by 3:
$$
f(2) = 3 \times f(3) = 3 \times 9 = 27.
$$
2. Next, express $f(2)$ in terms of $f(1)$:
$$
f(2) = \frac{1}{3} f(1).
$$
Again, solving for $f(1)$, multiply both sides by 3:
$$
f(1) = 3 \times f(2) = 3 \times 27 = 81.
$$
Thus, the value of $f(1)$ is $\boxed{81}$.
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