Another Nice Functional Equation

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[@shmuelzehavi4940]

Nice solution.
Just 2 little remarks:
1. It could be a little bit more efficient (at 1:46) to substitute: x = y + c.
2. Why the restriction to positive real numbers and not to R\{0} or even to C\{0} ?

[@sureshkrishna3787]

Great👍

[@ernstboyd8745]

this could be done with undetermined constants by guessing some form
for example f(x) = c1 + c2x + c3/x +... and finding the missing numbers

[@mystychief]

y=0 in the first equation gives f(f(x))=x. So twice the function of x gives back x, which gave me the idea of f(x)=1/x. When applied in the first equation it gave the right answer.

[@Attitude_boy_300]

Thumbnail reminds of emotional song 😂😂

[@zachariastsampasidis8880]

You don't know the value of c during your shift so f(y) is only determined for y bigger than c.
Edit: for clarity's sake, you can only substitute y=1 iff c

[@victorferreira5852]

Why x = 0 is not allowed? Doesnt seem obvious to me lol

[@doctorb9264]

Defined that way.

[@quantumcaffeine]

If you set x = 0, the RHS is zero for all y, meaning the only solution is the trivial f(x) = 0. So you need to exclude zero to get an interesting problem.

[@shmuelzehavi4940]

@@quantumcaffeine And this solution does not satisfy the original functional equation for
x, y ≠ 0

[@nalapurraghavendrarao6324]

Because set of +integers doesn't not include 0. Zero is neither + or -

[@bobbyheffley4955]

Division by zero

[@Thewerwolf]

why do you claim f(1) is constant

[@shmuelzehavi4940]

Because 1 is constant.

[@yoav613]

Noice