Question

x,y≥0 f(xf(y))f(y)=f(x+y) f(2)=0 f(x)≠0 0≤x<2 find f(1)

Answer 1

I wanna say f(1)=-1

Answer 2

why? obviously the function is not negative try x=1 y=1 then conclude f(f(1))f(1)=0

Answer 3

hmm does (xf(y))= x+y

Answer 4

now i wanna say 1

Answer 5

notice that for any \(z\ge 2\) we have

\[f(z)=f(v+2)\]\(v\ge 0\)

\[f(v+2)=f(vf(2))f(2)=f(vf(2))\times0=0\]

thus \(f(z)=0\) for all \(z\ge 2\)

then \[0=f(2)=f(1+1)=f(1f(1))=f(f(1))\]

so (provided \(f\ge0\)) \(f(1)\ge 2\)

Answer 6

small typo

\[0=f(2)=f(1+1)=f(1f(1))f(1)=f(f(1))f(1)\]
since \(f(1)\ne0\) we have \(f(f(1))=0\)

Answer 7

ok but what is the value of f(1)?