find all the functions $f:\mathbb{R}→\mathbb{R}$ such that $f(1)=1$ and $f(xy+f(x))=xf(y)+f(x)$

Question

here is what i did i found that $f(f(0))=f(0)$

i need to prove that f is injective so that $f(0)$ will be equal to $0$

Hence $f(f(x))=xf(0)+f(x)$ so if $f(0)=0$ and $f$ is injective $f(x)=x$ please don't give me the solution , just give me a hint to how to prove that $f$ is injective

Also , i found that $$f(x+f(x))=x+f(x)$$ i don't if that relation means that $f$ is injective

SORRY , GUYS! I FOUND A SOLUTION WITHOUT INJECTIVITY

N. S. · Accepted Answer · 2018-10-10T20:52:30.070

3

$x=1$ and $y=0$ yields

$$f(f(1))=f(0)+f(1)$$

Use $f(1)=1$.

Added Setting $y=0$ you get $$f(f(x))=f(x)$$

Next, set $y=\frac{1}{x}$ and use $f(f(x))=f(x)$ and $f(x+1)=f(x)+1$.

edited Oct 10 '18 at 20:52

answered Oct 10 '18 at 20:05

N. S.

132,525

1

I think this still doesn't address injectivity does it? – DanielV Oct 10 '18 at 20:08
yeah but now i have $f(x)=f(f(x))$ so i have to prove injectivity – user600785 Oct 10 '18 at 20:11
1

and i also have $f(y+1)=f(y)+1$ – user600785 Oct 10 '18 at 20:13
@user600785 Check the addition. – N. S. Oct 10 '18 at 20:52

find all the functions $f:\mathbb{R}→\mathbb{R}$ such that $f(1)=1$ and $f(xy+f(x))=xf(y)+f(x)$

1 Answers1