A Nice Homemade Functional Equation

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[@lamgiang5431]

Put y=0:f(x)=2x

[@tixanthrope]

You need to plug it back in the formula to check if it is well-defined, which it is.

[@theelk801]

if you do x=x and y=0 you get the answer immediately

[@wolliwolfsen291]

Why so complicated? The solution is obvious!

[@honestadministrator]

at y = 0 , f ( x) = 2 x This satisfies the given expression

[@DonEnsley-yi2ql]

f(x+yf(x)) = 2(x+y2x) f(z) = 2z One is a doubling function

[@iabervon]

If it's true for all y, it's true for y=0, and f(x)=2x. But it's worth checking that this also gives consistent results for non-zero y, in case y the answer is that no function works. However, 2(x+y(2x))=2x+4xy, as needed.

[@mystychief]

It looks linear. Fill in f(x) = ax+b (twice) then b must be zero and a must be 2.

[@leickrobinson5186]

Or just set y=0. Done.

[@bkkboy-cm3eb]

y=0 → f(x) = 2x x+yf(x) = x + 2xy f(x+yf(x)) = f(x + 2xy) = 2x +4xy ∴f(x) = 2x