Question

Find all functions f:R→R, that satisfy
f(x^3 +y^3) = x^2 f(x) + y f(y^2) for all x,y∈R

Answer 1

f(x) = x ---- 1

Answer 2

how to find all the functions?

Answer 3

Seems somewhat like linear function ... though not exactly!!

Answer 4

HI,
Let\[ f(x)=x^n. Then f(x^3 +y^3) = x^2 f(x) + y f(y^2)\] is not satisfied. So f(x) can't be a polynomial.
\[f(x).f(x)=x^2.y^2. NOw a = a^2 \iff a= 0 or 1\]

Answer 5

@experimentX I am confused pls help!

Answer 6

\[F(x+y)=F(x)+F(y)\]
put \[y=0\]
\[F(x)=F(x)+F(0)\]
so either\[ F(x)=0 \]
or \[F(0)=0 => F(x) \]has the form \[F(x)=xg(x)\] where \[g(x)=!0 \]for all \[x\]
now I have to show g(x)=1 for all x.

From the relations of \[F(x) and g(x)\], it can be derived
\[g(x+y)=(xg(x)+yg(y))/(x+y)\] and \[g(xy)=g(x)g(y)\]
put \[y=0\] in second relation
\[g(x)=g(x).g(0)\]
Rest @experimentX
& @Davidc Do IT ! YOu CAn

Answer 7

and the final answer is\[F(x) = x or F(x) = 0\]

Answer 8

A medal @Davidc + @experimentX pls....

Answer 9

This problem sucked me A LOT!

Answer 10

the final answer is f(x)=0 ??

Answer 11

ya!

Answer 12

but f(x)=x is true

Answer 13

i WROTE f(x)=0 OR f(x)=x

Answer 14

f(x) = 0 also satisfies the given condition .... though i can't think of any other right now!!

Answer 15

A medal!

Answer 16

i have a idea.
f(X)=k(x) , where k in R

Answer 17

mention anathor question!

Answer 18

Sorry Mention in other question!

Answer 19

At any nonzero multiple of pi.

--Elucidus

Answer 20

f(x^3 +y^3) = x^2 f(x) + y f(y^2)
f(x^3) = x^2 f(x)
f(y^3) = yf(y^2)

so x^2f(x) = xf(x^2) --> f(x^2) = xf(x) <---- not sure if i can do this!!

Answer 21

Oh man Tazzz not right!

Answer 22

So, f(x^3 +y^3)=f(x^3)+f(y^3)

Answer 23

That a linear function ...

Answer 24

f(a+b) = f(a) + (b) ...
but f(a^2) = af(a) is different .. (a is a variable)

Answer 25

a=x+1,
f((x+1)^2)=(x+1)f(x+1).......(1)

f((x+1)^2)=f(x^2+x+x+1)=f(x^2)+f(x)+f(x)+f(1)......(2)

Answer 26

how can you say the function is linear??

Answer 27

you say,
f(x^3 +y^3) = x^2 f(x) + y f(y^2)
f(x^3) = x^2 f(x)
f(y^3) = yf(y^2)

f(x^3 +y^3)=f(x^3)+f(y^3)
f(x+y)=f(x)+f(y)

Answer 28

Oh ... i got those two values putting x=0, y=0 simultaneously ...
f(x^3) = x^2 f(x)
f(y^3) = yf(y^2)

on these grounds i can say, xf(x) = f(x^2) .... but how did you conclude that??

Answer 29

f(x^3 +y^3) = x^2 f(x) + y f(y^2) =f(x^3)+f(y^3)

Answer 30

\( f(x^3) = x^{3/2} f(x^{3/2})\)

Answer 31

i don't understand

Answer 32

f(x^2) = xf(x)
not xf(x^2) = f(x^3)

Answer 33

f(x^3 +y^3) = x^2 f(x) + y f(y^2)
put y=0,
f(x^3) = x^2 f(x)

Answer 34

I don't think we could go back in reverse direction ...
f(x^3+y^3) = f(x^3) = f(y^3) <---- for x=0, y=0 <--- it doesn't mean f(x^3+y^3) = f(x^3)+f(y^3)

it's true only if x=0 or y=0

Answer 35

f(x^3+y^3) = f(x^3) = f(y^3) is not true , but
f(x^3) = x^2 f(x) for all x.
f(y^3) = yf(y^2) for all y

Answer 36

f(x^3 +y^3) = x^2 f(x) + y f(y^2)
when we put y=0, but x can be all real numbers
so it is true that f(x^3) = x^2 f(x) for all x.

Answer 37

yes it must be !! so for y

Answer 38

also,
f(x^3 +y^3) = x^2 f(x) + y f(y^2)
put x=0, but y can be all real numbers
so it is true that f(y^3) = yf(y^2) for all y
Thus, f satisfy f(x^3) = x^2 f(x) for all x and f(y^3) = yf(y^2) for all y

Answer 39

since both x and y are domain,
x^2f(x) = xf(x) can be said!!

Answer 40

*x^2f(x) = xf(x^2)

Answer 41

LOL ... sorry .. i posted on wrong section

Answer 42

Never mind

Answer 43

It is easy to show that f(x)= a x does satisfy the condition. The question is to see if the converse is true.

Answer 44

Let
\[
c= x^{\frac 1 3}\\
d=x^{\frac 1 3}\\
f(x+y)= f(c^3 + d^3) = f(c^3) + f(d^3)= f(x)+ f(y)
\]
If we know in addition that f is continuous, then
\[f(x)= a \ x
\]