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f-R-R-If-x-2-f-x-f-1-x-2x-x-4-Determine-f-x-

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Question Number 11517 by Joel576 last updated on 27/Mar/17
f : R → R If x^2 f(x) + f(1−x) = 2x − x^4 Determine f(x)
f:R→RIfx2f(x)+f(1−x)=2x−x4Determinef(x)
Answered by sma3l2996 last updated on 27/Mar/17
let u=1−x (1−u)^2 f(1−u)+f(u)=2(1−u)−(1−u)^4 so f(1−u)=((2−(1−u)^3 )/((1−u)))−((f(u))/((1−u)^2 )) f(1−x)=((2−(1−x)^3 )/((1−x)))−((f(x))/((1−x)^2 )) x^2 f(x)+f(1−x)=x^2 f(x)−((f(x))/((1−x)^2 ))+((2−(1−x)^3 )/((1−x))) f(x)(((x^2 (1−x)^2 −1)/((1−x)^2 )))+((2−(1−x)^3 )/((1−x)))=2x−x^4 f(x)=(2x−x^4 +(((1−x)^3 −2)/((1−x))))((((1−x)^2 )/(x^2 (1−x)^2 −1)))
letu=1−x(1−u)2f(1−u)+f(u)=2(1−u)−(1−u)4sof(1−u)=2−(1−u)3(1−u)−f(u)(1−u)2f(1−x)=2−(1−x)3(1−x)−f(x)(1−x)2x2f(x)+f(1−x)=x2f(x)−f(x)(1−x)2+2−(1−x)3(1−x)f(x)(x2(1−x)2−1(1−x)2)+2−(1−x)3(1−x)=2x−x4f(x)=(2x−x4+(1−x)3−2(1−x))((1−x)2x2(1−x)2−1)
Commented by Joel576 last updated on 28/Mar/17
thank you very much
thankyouverymuch
Commented by Joel576 last updated on 28/Mar/17
one more, how can we determine the value of f(2009) ?
onemore,howcanwedeterminethevalueoff(2009)?
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/Mar/17
x^2 f(x)+f(1−x)=2x−x^4 (1−x)^2 f(1−x)+f(x)=2(1−x)−(1−x)^4 [x^2 (1−x)^2 −1]f(x)=(2x−x^4 )(1−x)^2 −2(1−x)+(1−x)^4 = (1−x)[2x−x^4 −2x^2 +x^5 −2+1−3x+3x^2 −x^3 )= (1−x)(x^5 −x^4 −x^3 +x^2 −x−1)= −x^6 +x^5 +x^4 −x^3 +x^2 +x+x^5 −x^4 −x^3 +x^2 −x−1= −(x^6 −2x^5 +2x^3 −2x^2 +1)= −(x^3 −x^2 +1)^2 +x^4 =(x^2 +x^3 −x^2 +1)(x^2 −x^3 +x^2 −1)= (x^3 +1)(−x^3 +2x^2 −1)=(x^3 +1)(−x(x^2 −2x+1)−(1−x))= (x^3 +1)(−x(1−x)^2 −(1−x)) ⇒f(x)=(((1+x)(1−x+x^2 )(1−x)(1+x(1−x)))/((1−x(1−x))(1+x(1−x))))= f(x)=(1−x)(1+x)=1−x^2 f(2009)=(1−2009)(1+2009)= −2010×2008
x2f(x)+f(1−x)=2x−x4(1−x)2f(1−x)+f(x)=2(1−x)−(1−x)4[x2(1−x)2−1]f(x)=(2x−x4)(1−x)2−2(1−x)+(1−x)4=(1−x)[2x−x4−2x2+x5−2+1−3x+3x2−x3)=(1−x)(x5−x4−x3+x2−x−1)=−x6+x5+x4−x3+x2+x+x5−x4−x3+x2−x−1=−(x6−2x5+2x3−2x2+1)=−(x3−x2+1)2+x4=(x2+x3−x2+1)(x2−x3+x2−1)=(x3+1)(−x3+2x2−1)=(x3+1)(−x(x2−2x+1)−(1−x))=(x3+1)(−x(1−x)2−(1−x))⇒f(x)=(1+x)(1−x+x2)(1−x)(1+x(1−x))(1−x(1−x))(1+x(1−x))=f(x)=(1−x)(1+x)=1−x2f(2009)=(1−2009)(1+2009)=−2010×2008

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