Question

A little bored so i am going to ask a QH question :D

Answer 1

The function F satisfies

\[(x-y)f(x+y)-(x+y)f(x-y)+4xy(x ^{2}-y ^{2})=0\]

whenever \[\left| x \right|\neq \left| y \right|\]

Given that f(1)=2. what is
f(x)

Answer 2

@dan815 @ganeshie8

Answer 3

Anybody? 12mins have gone by D:

Answer 4

need help

Answer 5

Of course

Answer 6

i need help

Answer 7

Its an even function.

Answer 8

?? what

Answer 9

you have any choices??

Its an even function,
because when you plug in x=0, y not =0,
you get f(y) = f(-y)

Answer 10

i had choices lol

this is off this site dan gave me and i had OB so i just asked it lol

Answer 11

can f(x)=2x?

Answer 12

but i have the answer i just need you guys the figure it out :D

sorta like a mini riddle or whatever

Answer 13

i have no idea beside f(x)=2x :D

Answer 14

Nope thats not the answer :/

Answer 15

I'd definitely substitute the elliptic substitution lol.

a=x+y, b=x-y

Answer 16

This will give you:
\[b f(a) - a f(b)+(a^2-b^2)ab=0 \] At least that's nicer to look at don't you think so?

Answer 17

\[\text{ If } x+y=1 \text{ then } y=-x+1 \text{ and so } x-y=2x-1 \\ \text{ so we have } \\ (2x-1)f(1)-f(2x-1)+4x(-x+1)(2x-1)=0 \\ (2x-1)(2)-f(2x-1)+4x(-x+1)(2x-1)=0 \\ f(2x-1)=\frac{4x(-x+1)(2x-1)}{2(2x-1)}\]

Answer 18

\[u=2x-1 \\ \text{ then } x=\frac{u+1}{2}\]

Answer 19

\[f(u)=\frac{4 \cdot \frac{u+1}{2}(-\frac{u+1}{2}+1)(u)}{2u} \text{ assuming } 2x-1=u \neq 0 \\ \text{ we have } \\ f(u)=2 \frac{u+1}{2}(\frac{-u-1}{2}+1)\]

Answer 20

\[f(u)=(u+1)(\frac{-u-1+2}{2}) \\ f(u)=(u+1)(\frac{-u+1}{2})\]

Answer 21

\[f(u)=\frac{1}{2}(1+u)(1-u) \\ f(u)=\frac{1}{2}(1-u^2)\]

Answer 22

I think I made a mistake somewhere since f(1) doesn't equal 2 here