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Question Number 117518 by bobhans last updated on 12/Oct/20
f(x+1)+xf(1−x)=x^2 f^(−1) (x) =?
f(x+1)+xf(1−x)=x2f−1(x)=?
Answered by bemath last updated on 12/Oct/20
replace x+1 by x (ii) f(x)+(x−1)f(1−(x−1))=(x−1)^2 ⇒ f(x)+(x−1)f(2−x)=(x−1)^2 (iii) replace 1−x by x ⇒f(1−x+1)+(1−x)f(x)=(1−x)^2 f(2−x)+(1−x)f(x)=(x−1)^2 multiply eq (iii) by (x−1) ⇒f(x)+(x−1)f(2−x)=(x−1)^2 ⇒−(1−x)^2 f(x)+(x−1)f(2−x)=(x−1)^3 _______________________________ − ⇒ (1+(1−x)^2 ) f(x) = (x−1)^2 −(x−1)^3 ⇒(2−2x+x^2 ) f(x) = (x−1)^2 (1−(x−1)) f(x) = (((x−1)^2 (2−x))/(x^2 −2x+2))
replacex+1byx(ii)f(x)+(x−1)f(1−(x−1))=(x−1)2⇒f(x)+(x−1)f(2−x)=(x−1)2(iii)replace1−xbyx⇒f(1−x+1)+(1−x)f(x)=(1−x)2f(2−x)+(1−x)f(x)=(x−1)2multiplyeq(iii)by(x−1)⇒f(x)+(x−1)f(2−x)=(x−1)2⇒−(1−x)2f(x)+(x−1)f(2−x)=(x−1)3_______________________________−⇒(1+(1−x)2)f(x)=(x−1)2−(x−1)3⇒(2−2x+x2)f(x)=(x−1)2(1−(x−1))f(x)=(x−1)2(2−x)x2−2x+2

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