I Solved A Homemade Functional Equation

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[@ryanstaal3233]

Solved it in a few seconds. Just substitute x=-f(y)/2 and you get that f(y)=4y+c for a constant c. Plugging this in gives that only c=0 works.

[@SyberMath]

Wow! That was quick

[@scottleung9587]

Nice job!

[@SyberMath]

Thank you! Cheers!

[@JohnSmith-mz7dh]

There is another way. I let y=0 and differentiated it once on both sides. 2f’(2x+f(0)=8. f’(2x+f(0))=4. We can let some variable t=2x+f(0), so f’(t)=4. f(t)=4t+c. We can immediately see that c=0 since there is no constant term on the right hand side.
I think it’s obvious that f is differentiable. We know that f’(t)=4, no possibility for being undefined! I’m not sure exactly how to prove this more rigourisly however.

[@srirajan1933]

You are great, @SyberMath! You create compelling content for us each day!! You remind me of my high school math teacher, Daniel Dawson, who inspired so many of us in the late 1970s to learn mathematical thinking. Quick thought: if you replace f(2x + f(y)) on LHS with f(2x + 2f(y)), or more exotic with powers and logs, such as f(2x + f^2(y)) so that your substitution method does *not* cancel f(y) on LHS to isolate f(y) by itself, this gets tricky... will give it a try and see where this goes. Love your work, buddy! Thank you immensely for helping educate (or re-educate) all of us!!

[@SyberMath]

Thank you for the kind words! ❤️

[@mystychief]

It looks as if f(x) is linear, so suppose f(x) = ax + b. There is twice an f in the first equation, so twice fill in f(x) = ax + b, you get 2ax+af(y)+b = 2ax + a²y + ab + b = 8x + 16y. This gives 2a=8 or a=4 and b=0. So f(x) = 4x (a²y=16y also fits).

[@florianbuerzle2703]

Solved it also in seconds: substitute x = x/2 - f(y)/2, then you will get f(x) + 4f(y) = 4x +16y which is solved by f(x) = 4x 😊

[@andirijal9033]

f(y) = -x so y = inverse of (-x) and f(x) = ax + b

[@kianmath71]

F(x) = 4x

[@giuseppemalaguti435]

f(x)=4x

[@alextang4688]

Put x=0, y=0, get f(f(0))=0
Put x=0, y=f(0), get f(0)=0
Put x=2a, y=0, get f(a)=4a answer. 😉😉😉😉😉😉