A Nice and Easy Functional Equation

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[mystychief]

The contradiction in this functional equation is always there (independent of the premise that f(x) it is linear or filling in certain values). Suppose t=1-x, so x=1-t. Filling this in the first functional equation gives f(1-t)-f(t)=1-t or f(1-x)-f(x)=1-x. Substract this from the first functional equation and you get 0=1. This contradiction proves that the first functional equation can never be possible for any x and there is no function possible for this functional equation.

[paulortega]

Yep, let replace x with 1-x and you get f(x) - f(1-x) = x and f(x) - f(1-x) = x -1 ouch

[brendanward]

f(x) - f(1-x) = f(x)
f(1-x) - f(1-(1-x) = f(1-x) => f(1-x) - f(x) = f(1-x)
Add both equations:
0 = 1 => no solution.

[nanamacapagal]

ATTEMPT:

There's a symmetry in the equation. Plugging in y = 1 - x gives
f(1-y) - f(y) = 1-y
Multiply by -1 and switch out the dummy variable:
f(x) - f(1-x) = x - 1
But it was stated earlier that f(x) - f(1-x) = x.
So x = x - 1, and that's a contradiction.
There are no solutions to the functional equation.

[rajeshbuya]

I'm so very interested in your Functional equation-related problems.

Can you please help solve
f ' (x) = f (x + 1)

[StuartSimon]

Can you please explain how you come up with functional equations with no solution? Branching out from a contradiction to me seems harder than branching out from a solution.

[williamspostoronnim]

Неужели это так обязательно - трещать как сорока?