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Function

Let f(x) be a function such that f(ab) = bf(a) + af(b) for all nonzero real numbers. Given that f(4) = 3, which of the following is a possible value of f(2018)? (A) 0 (B) 3/4 (C) 4/3 (D) 1512 (E) 2688

By substitution, it's easy to find that f(1) = 0 and f(2) = 3/4. But how can I get to f(2018)?
Original post by Prasiortle
Let f(x) be a function such that f(ab) = bf(a) + af(b) for all nonzero real numbers. Given that f(4) = 3, which of the following is a possible value of f(2018)? (A) 0 (B) 3/4 (C) 4/3 (D) 1512 (E) 2688

By substitution, it's easy to find that f(1) = 0 and f(2) = 3/4. But how can I get to f(2018)?


Can't see a solution myself. A bit of playing produces some properties, such as:

f(1/4)=f(1/2)f(1/4) = f(1/2) so it's not an injective function.

Also, f(an)=nan−1f(a)f(a^{n}) =na^{n-1}f(a)

and f(1/a)=−1a2f(a)f(1/a) = -\frac{1}{a^2}f(a)

Makes me think there's something to do with differentiation in there, but no idea if that will/can lead anywhere.

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