## 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 $$f(x)$$.

1. ganeshie8

let $$y = 0$$, $f(x)f(0)-f(0) = x\tag{1}$ 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 $$f(x) = x+1$$

2. freckles

I was just fixing to ask why :p

3. freckles

I also noticed I get two possibilities for f(1)

4. freckles

I don't know how to know which one is the right one for f(1)

5. thomas5267

\text{Assume }f(0)=0\\ \begin{align*} f(0)f(x)-f(0)&=x\\ 0&=x \end{align*}

6. freckles

$f^2(1)-f(1)-2=0 \\ (f(1)-2)(f(1)+1)=0 \\ f(1)=2 \text{ or } f(1)=-1$

7. thomas5267

I also did a similar thing then ganeshie8 posted his answer.

8. ganeshie8

just 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..

9. thomas5267

\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*}

10. freckles

ok cool and if f(1)=2 $f(1)f(x)-f(x)=x+1 \\ 2f(x)-f(x)=x+1 \\ f(x)=x+1 \\ \text{ checking } \\ f(x)f(y)-f(xy)=(x+1)(y+1)-(xy+1) \\ = xy+x+y+1-xy-1 \\ =x+y$ great stuff @thomas5267

11. thomas5267

One 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.

12. beginnersmind

@zmudz can you give us some background on these problems? Are you looking for hints, solution methods or the solution?

13. anonymous

i assumed its one to one and onto function :3 f(x) f(y) - f(xy) = x + y yx-xy=f(y)+f(x) xy-xy=f(x)+f(y) 0=f(x)+f(y) f(x)+f(y)=0 f(x)=-f(y) so f(x)=-x satisfy :)

14. freckles

I can't figure out how you got any of that. But checking your solution... we see that $-x \cdot (-y)-(-xy) \neq x+y$

15. anonymous

hmmm you are right