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Algebra Difficulty 4.8 AIME Prove it Slovenia

Find all functions f:R→Rf: \mathbb{R} \to \mathbb{R}, such that
f(xy)=xf(y)+3f(x)+3 f(xy) = x f(y) + 3 f(x) + 3
for all x,y∈Rx, y \in \mathbb{R}.

Solution

Plugging x=0x = 0 into the equation, we get f(0)=3f(0)+3f(0) = 3 f(0) + 3, which implies that f(0)=−32f(0) = -\frac{3}{2}.

Now plug in y=0y = 0 to get f(0)=xf(0)+3f(x)+3f(0) = x f(0) + 3 f(x) + 3. From here we can obtain
f(x)=−13xf(0)+13f(0)−1=12x−32. f(x) = -\frac{1}{3} x f(0) + \frac{1}{3} f(0) - 1 = \frac{1}{2} x - \frac{3}{2}.
It is easy to verify that the function f(x)=12x−32f(x) = \frac{1}{2} x - \frac{3}{2} satisfies the given equation, and this is the only solution.

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