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

2 Answers

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)

Jun 5, 2017

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