Question

The function `f : RR -> RR` satisfies `xf(x) + f(1 - x) = x^3 - x` for all real `x`. Find `f(x)`?

Answer 1

`x(x - 1) = f(x)`

Explanation:

`f(1-x) = x^3 - x - x f(x)`

Exchange `x` by `1-x`.

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

`f(1-x) = (1-x)^2 - 1 - frac{f(x)}{1 - x}`

Equals to the first:

`x^3 - x - x f(x) = (1 -2x + x^2) - 1 + frac{f(x)}{x - 1}`

`x^3 - x^2 + x = x f(x) + frac{f(x)}{x - 1}`

`x(x^2 - x + 1)/(x + 1/(x-1)) = f(x) = x(x-1)(x^2 - x + 1)/(x^2 - x + 1)`

Answer 2

`f(x)=x^2-x`

Explanation:

Making `f(x) = a x^2+b x +c` and applying in

`x f(x)+f(1-x)=x^3-x` we have

`x(a x^2+b x +c)+a (1-x)^2+b(1-x)+c = x^3-x` and grouping coefficients

`(a-1)x^3+(a+b)x^2+(1-2a-b+c)x+a+b+c=0`

this must be verified for all `x` so

`{(a-1=0),(a+b=0),(1-2a-b+c=0),(a+b+c=0):}`

solving we get

`a=1,b=-1,c=0` so

`f(x)=x^2-x`