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zmudz
 one year ago
The function \(f : \mathbb{R} \rightarrow \mathbb{R}\) satisfies \(f(x) f(y)  f(xy) = x + y\) for all \(x\), \(y \in \mathbb{R}\). Find \(f(x)\).
zmudz
 one year ago
The function \(f : \mathbb{R} \rightarrow \mathbb{R}\) satisfies \(f(x) f(y)  f(xy) = x + y\) for all \(x\), \(y \in \mathbb{R}\). Find \(f(x)\).

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4let \(y = 0\), \[f(x)f(0)f(0) = x\tag{1}\] let \(x=0\) in above equation : \[f(0)f(0)f(0) = 0\implies f(0) = 0\lor 1\] however \(f(0)=0\) is not possible (why?) therefore \(f(0)=1\) plug that back in equation \((1)\) to get \(f(x) = x+1\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I was just fixing to ask why :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I also noticed I get two possibilities for f(1)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I don't know how to know which one is the right one for f(1)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \text{Assume }f(0)=0\\ \begin{align*} f(0)f(x)f(0)&=x\\ 0&=x \end{align*} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f^2(1)f(1)2=0 \\ (f(1)2)(f(1)+1)=0 \\ f(1)=2 \text{ or } f(1)=1\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I also did a similar thing then ganeshie8 posted his answer.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4just an example to convince myself on why there are false values : \(f = 1 \implies f^2 = 1 \implies f = \pm 1 \) ofcourse \(f=1\) is extraneous. since we're squaring the function, we are increasing the degree and thus increasing the number of solutions, so we must check for extraneous stuff in the end..

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \text{Assume }f(1)=1\\ \begin{align*} f(1)f(x)f(x)&=x+1\\ 2f(x)&=x+1\\ f(x)&=\frac{x+1}{2}\\ f(x)f(y)f(xy)&=\frac{x+1}{2}\frac{y+1}{2}\frac{xy+1}{2}\\ &=\frac{xy+x+y+1}{4}\frac{xy+1}{2}\\ &\neq x+y \end{align*} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok cool and if f(1)=2 \[f(1)f(x)f(x)=x+1 \\ 2f(x)f(x)=x+1 \\ f(x)=x+1 \\ \text{ checking } \\ f(x)f(y)f(xy)=(x+1)(y+1)(xy+1) \\ = xy+x+y+1xy1 \\ =x+y\] great stuff @thomas5267

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1One of the question on the only mathematics competition I have participated is to find the value of \(f(0)\) with \(f(x)f(y)=f(x+y)\). It can be solved with a similar method.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0@zmudz can you give us some background on these problems? Are you looking for hints, solution methods or the solution?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i assumed its one to one and onto function :3 f(x) f(y)  f(xy) = x + y yxxy=f(y)+f(x) xyxy=f(x)+f(y) 0=f(x)+f(y) f(x)+f(y)=0 f(x)=f(y) so f(x)=x satisfy :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I can't figure out how you got any of that. But checking your solution... we see that \[x \cdot (y)(xy) \neq x+y\]
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