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Question Number 3653 by prakash jain last updated on 17/Dec/15
f:R→R f(x−f(y))=f(f(y))+xf(y)+f(x)−1 f(x)=?
f:R→Rf(x−f(y))=f(f(y))+xf(y)+f(x)−1f(x)=?
Commented by Rasheed Soomro last updated on 18/Dec/15
f(x−f(y))=f(f(y))+xf(y)+f(x)−1................(i) −−−−−− x=0 in (i) f(0−f(y))=f(f(y))+0f(y)+f(0)−1 f( −f(y) )=f( f(y) )+f(0)−1 −−−−−−− y=0 in (i) f(x−f(0))=f(f(0))+xf(0)+f(x)−1 Let y=x f(x−f(x))=f(f(x))+xf(x)+f(x)−1 f(x−f(x))=f(f(x))+(x+1)f(x)−1 f(x−f(x))−f(f(x))+1=(x+1)f(x) x↔y in (i) f(y−f(x))=f(f(x))+yf(x)+f(y)−1................(ii)
f(x−f(y))=f(f(y))+xf(y)+f(x)−1…………….(i)−−−−−−x=0in(i)f(0−f(y))=f(f(y))+0f(y)+f(0)−1f(−f(y))=f(f(y))+f(0)−1−−−−−−−y=0in(i)f(x−f(0))=f(f(0))+xf(0)+f(x)−1Lety=xf(x−f(x))=f(f(x))+xf(x)+f(x)−1f(x−f(x))=f(f(x))+(x+1)f(x)−1f(x−f(x))−f(f(x))+1=(x+1)f(x)x↔yin(i)f(y−f(x))=f(f(x))+yf(x)+f(y)−1…………….(ii)
Commented by prakash jain last updated on 18/Dec/15
f(y)=z f(x−z)=f(z)+xz+f(x)−1 ...(i) assume f(x)=0 for some x then we can try z=0 z=0, x=0 f(x)=f(0)+f(x)−1 f(0)=1 put x=0 f(−z)=f(z) even function continue
f(y)=zf(x−z)=f(z)+xz+f(x)−1…(i)assumef(x)=0forsomexthenwecantryz=0z=0,x=0f(x)=f(0)+f(x)−1f(0)=1putx=0f(−z)=f(z)evenfunctioncontinue

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