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

Is this function only defined for x between 0 and 2?

Answer 2

no [0,inf]

Answer 3

look at the first condition

Answer 4

Ah, alright.

Answer 5

I've gotten that f(0) = 1 by letting both x = y = 0, but not much beyond that.

Answer 6

yeap but i am looking for f(1) ok go ahead.

Answer 7

I see that f(f(1) = 0, so f(1) could be 2, but it could be other things as well.

Answer 8

I'm going to go out on a limb, and say that \(f(1)=2\).

Answer 9

f(f(1))*

Answer 10

If we let x=y=1, then we have that \(f(1\cdot f(1))=f(2)\). Thus, \(1\cdot f(1)=2\). It follows that \(f(1)=2\).

Answer 11

I did that wrong. one moment.

Answer 12

yeah, I just got that too. f(x) = 0 for all x greater than 2.

Answer 13

0
no cause of x=1 y=2 f(3)=0
you can see also f(x)=0 for all x>=2

Answer 14

We have \(f(1\cdot f(1))\cdot f(1)=f(2)=0\) So either \(f(1\cdot f(1))=0\) or \(f(1)=0\).

Since we're told \(f(x)\neq 0\), it must be that \(f(1\cdot f(1))=0\)

Answer 15

\(f(x)\neq 0\) for \(0\leq x <2\)

Answer 16

ok then f(1)>=2

Answer 17

what is exactly?

Answer 18

If you let x = 1 and y = f(1), then you get:\[f(1\cdot f(f(1))f(f(1))=f(1+f(1))\Longrightarrow f(f(f(1)))f(f(1))=f(1+f(1))\]Since f(f(1))=0, we have:\[f(f(1)+1)=0\Longrightarrow f(1)+1\ge 2\Longrightarrow 1\le f(1)\le 2\]getting close >.<

Answer 19

No wait, now i did something wrong >.< one sec.

Answer 20

it just has to be bigger than 1, not less that 2, scratch that.

Answer 21

only >=1 we knew that!

Answer 22

>.< haha

Answer 23

certainly you have wonderful insight!

Answer 24

Just trying random things, but nothing is creating a solution >.<

Answer 25

Do you know if there's supposed to be a specific number?

Answer 26

i think somebody who immediately understand wrong way is successful.dont you?

Answer 27

no there is not.

Answer 28

some help! try with x=2/f(1)

Answer 29

I'm sorry, but I've got nothing else on this problem, and I've got to go. Good luck.

Answer 30

good luck

Answer 31

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

f(0f(1))f(1)=f(1)
f(0)f(1)=f(1)
f(0)=1

Great I am not going anywhere..... Let me try more