Let c=f(1). If we plug x=y=1 into the functional equation we get f(c) = f(f(1)) = 4. Plugging in x=0 (regardless of the value of y) we get f(0)=0. From here we see that c is not 0. Now plug in y=1. We have f(cx) = 4x. Through substitution f(x) = 4x/c. So c = f(1) = 4/c, and thus c^2 = 4. We thus have c=2 or c=-2. Using f(x) = 4x/c, these respectively give the solutions f(x) = 2x and f(x) = -2x. It’s easy to check that these are both solutions to the original functional equation.
A "sneaky" way is to substitute x=1/f(u) and y = u where f(u) != 0 . This quickly gives us f(u) = 4u/f(1) . Substituting u=1 gives us the choice of f(1) = ±2 , and so f(u) = ±2u . When f(u)=0, substitute y=u into the initial equation to arrive at f(0)=0, which is consistent with our f(u) = ±2u solutions.
from f(f(x)) = 4x we can see f is 1-1 so f(x(f(y)) = f(yf(x)) implies xf(y) = yf(x) so f(x)/x must be constant for nonzero x. together with f(0) = 0 we have linearity and the solution becomes obvious.
We differentiate with respect to x. f’(xf(y))f(y)=4y. Set x=0. f(y)=4y/f’(0) Assuming f’(0) is defined, then… f(y)=4y/c. Where c is some nonzero constant. Let x=1 and y = 1 in our original equation… f(f(1))= 4 so 16/c^2=4. c^2=4. c=+-2. f(x)=-2x,2x
f(1) = k f(xk) = 4x f((1/k)*k) = f(1) = k = 4/k; k = ± 2 1) k = 2 --> f(2x) = 4x --> f(n) = 2n 2) k = -2 --> f(-2x) = 4x --> f(n) = -2n Both are correct, and other can't be correct. So f(n) = ±2n
f(0) = 0. and f ( f (y)) = 4 y Hereby f(y) = a y + b as f (0) = 0, f ( y) = a y f ( f (y)) = a ( a y ) Hereby a^2 = 4 Either f (y) = 2 y Or f ( y) = - 2 y
Let c=f(1).
If we plug x=y=1 into the functional equation we get f(c) = f(f(1)) = 4. Plugging in x=0 (regardless of the value of y) we get f(0)=0. From here we see that c is not 0.
Now plug in y=1. We have f(cx) = 4x. Through substitution f(x) = 4x/c. So c = f(1) = 4/c, and thus c^2 = 4. We thus have c=2 or c=-2.
Using f(x) = 4x/c, these respectively give the solutions f(x) = 2x and f(x) = -2x. It’s easy to check that these are both solutions to the original functional equation.
very smart!
Cool!
A "sneaky" way is to substitute x=1/f(u) and y = u where f(u) != 0 . This quickly gives us f(u) = 4u/f(1) . Substituting u=1 gives us the choice of f(1) = ±2 , and so f(u) = ±2u .
When f(u)=0, substitute y=u into the initial equation to arrive at f(0)=0, which is consistent with our f(u) = ±2u solutions.
from f(f(x)) = 4x we can see f is 1-1
so f(x(f(y)) = f(yf(x)) implies xf(y) = yf(x)
so f(x)/x must be constant for nonzero x. together with f(0) = 0 we have linearity and the solution becomes obvious.
We can see that the null function is a solution
We assume thar f is non null and let yf(x)= t to have f(t)/t= 4 and we get the sol
We differentiate with respect to x. f’(xf(y))f(y)=4y.
Set x=0.
f(y)=4y/f’(0)
Assuming f’(0) is defined, then…
f(y)=4y/c. Where c is some nonzero constant.
Let x=1 and y = 1 in our original equation…
f(f(1))= 4
so 16/c^2=4. c^2=4. c=+-2.
f(x)=-2x,2x
once you have f(f(y))=4y you don’t have to assume that f is linear, it follows from what you already have
I don't get the reasoning, just because f(f(x)) is linear doesn't mean f(x) is. Counterexample is f(x)=1/x
I think you can rule out c=0 like this::
Suppose f(1)=0. Replace y with 1
f(xf(1)) = 4x for all x
f(0) = 4x for all x, a contradiction
Easy, f(x)=2x
숫자 아무거나 넣어 봅시다. 2의 제곱이 4니까 대충 2라고 추측해 보자고요~
f(x)=2x
=f(x f(y)) = f(2xy) = 4xy
쉽죠?
-2도 같죠? -2도 제곱하면 4니까?
f(x)=-2x
f(xf(y)) = f(-2xy) = 4xy
플러스 마이너스 다 되죠?
f(x)= 1 :+2x , 2 : -2x
f(1) = k
f(xk) = 4x
f((1/k)*k) = f(1) = k = 4/k; k = ± 2
1) k = 2 --> f(2x) = 4x --> f(n) = 2n
2) k = -2 --> f(-2x) = 4x --> f(n) = -2n
Both are correct, and other can't be correct. So f(n) = ±2n
Finally this proves that f =0 is also a solution. Just kidding. Compose by f twice to get the solution.
f(x) = 2x
👍👍👍
f( x ) = 2x , f( y ) = 2y
f( x f( y ) ) = f( x ( 2y ) ) = 2( x ( 2y ) ) =
= 4xy
f( x ) = - 2x , f( y ) = - 2y
f( x f( y ) ) = f( x ( - 2y ) ) =
= - 2 ( x ( - 2y ) ) =
= 4xy
😊🤪👍👋
f(0) = 0. and f ( f (y)) = 4 y
Hereby f(y) = a y + b
as f (0) = 0, f ( y) = a y
f ( f (y)) = a ( a y )
Hereby a^2 = 4
Either f (y) = 2 y
Or f ( y) = - 2 y