Question

Prove that there is no function `f` defined in `RR` for which it applies helpp?:(

`f(x+1)+f(1-x)=2x+3` , `∀ x in RR`

Answer 1

See explanation...

Explanation:

Given:

`f(x+1)+f(1-x) = 2x+3`

We find:

`1 = 2(color(blue)(-1))+3 = f((color(blue)(-1))+1) + f(1-(color(blue)(-1))) = f(0)+f(2)`

`= f(2)+f(0) = f((color(blue)(1))+1)+f(1-(color(blue)(1))) = 2(color(blue)(1))+3 = 5`

Which is false.

So there is no such function `f(x)` defined for all `x in RR`

Answer 2

See below.

Explanation:

Considering `x = -1` we have

`f(0)+f(2) = 1`

now considering `x = 1` we have

`f(2)+f(0) = 5`

So, no such function exists.