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

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Answer 1 · 2017-06-05T14:13:26

#x(x - 1) = f(x)#

#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 · 2017-06-05T14:17:34

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

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#