Find all real functions $f:R\mapsto R$ satisfying the relation   $$f(x^2+yf(x))=xf(x+y).$$   My Answer: Putting $y=0$ we get, $f(x^2)=xf(x)$. o  Which implies $\frac{f(x^2)}{x^2}=\frac{f(x)}{x}$.  Let $g(x)=\frac{f(x)}{x}$.(Assume $x\ne 0$)  Hence, $g(x)=g(x^2)$.  Therefore, $g(x)=c$, Or $f(x)=cx$ (when $x \ne 0$)  And $f(0)=0$ (By putting $x=y=0$).  Is my answer is ask right?? If not then kindly tell me the mistake!
85996
To find all functions $f$ which is a real function from $\Bbb R \to \Bbb R$ satisfying the relation  $$f(x^2 + yf(x)) = xf(x+y)$$  It can be easily seen that the identity function $i.e.$ $f(x)=x$ and $f(x)=0$ (verified just now) satisfies the above relation!! And putting $y=0$ I have got $f(x^2)=xf(x)$.  Help needed to find other functions satisfying the relation.
89311
Determine all $f : \mathbb R \rightarrow \mathbb R$ such that $$f\big(f(x) + y\big) = f(x^2 - y) + 4yf(x)$$ for all $x,y\in\mathbb{R}$.   $f(x) = 0$ is obvious , but how can we find all other functions?  Thanks in advance!
94317
Find $f:f(xf(x)+f(y))=f(x)^2+y$  Domain and co-domain is real numbers I did the following:  Let $s=f(0)$  Then $f(f(y))=s^2+y$ so $f$ is surjective Also, $f(x)=f(y)\implies f(xf(x)+f(y))=f(xf(x)+f(x))\implies x=y$ so $f$ is injective So, $f$ is bijective. Letting $f(x)=0, y=0$  we get $f(f(0))=0 \implies s=f(0)=0$  In fact $f(f(x))=x$ for ALL $x$ Letting $x=1,y=0$ we get  $f(f(1))=f(1)^2\implies 1=f(1)^2 \implies f(1)=1$ or  $f(1)=-1$ If $f(1)=1$ we let $x=1$ to get  $f(y+1)=f(y)+1$  By induction this leads to $f(x)=x$ for all integers If $f(1)=-1$ we let $x=1$ to get  $f(y-1)=f(y)+1$  By induction we get $f(x)=-x$ for all integers How to extend the domain over real numbers I don't know, any help would be appreciated
39788
If a Continuous function $f(x)$ satisfies $f(x) = f(2x)$, for all real $x$, then does $f(x)$ necessarily have to be constant function? If so, how do you prove it? If not any counter examples?
31188
Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ without any further restriction. If $f(f(x))=x^2-x+1$, how can one find $f(0)$?   Thanks in advance.
51298
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$  My attempt:   If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.   Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon&gt;0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$.   Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!
73757
Let $\Bbb R $ be the set of real numbers .Determine all functions $f:\Bbb R \rightarrow \Bbb R $ satisfying the equation   $$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$ for all real numbers $x$ and $y$.   My work  $$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$ Setting $y=0$ $$f(x + f(x)) + f(0) = x + f(x )$$ $$\implies f(x + f(x))  = x + f(x ) -f(0)$$ $$\implies f(t)  = t -f(0)$$ Setting $x=0$ $$f(f(y)) + f(0) = 0 + f(y) +yf(0)$$ $$f(f(y)) =  f(y) +f(0)(y-1)$$  Setting $y=1$ $$f(x + f(x + 1))  = x + f(x + 1)$$
94971
let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$     My Try :  $$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$  So we have :  $$f(x)=x^2-1$$  it is right ?Is there another answer?
95419
I'm getting $f(x)=2x+f(0)$ and $f(x)=f(0)-2x$ by setting $y=0$, but I'd like to verify. Am I right?
91399
let $f$ be a polynomial function such that $f(\mathbb{Q})\subset \mathbb{Q}$ and $f(\mathbb{R}-\mathbb{Q}) \subset \mathbb{R}- \mathbb{Q}$ . Prove that $f(x)=ax+b$ for some $a,b\in\mathbb{Q}$   My approach :  I started by defining $g(x)=f(x)-f(0)$ .   Now $g(x)$ satisfies the properties $g(\mathbb{Q})\subset \mathbb{Q}$ ,  $g(\mathbb{R}-\mathbb{Q})\subset\mathbb{R}-\mathbb{Q}$ and $g(0)=0$ . $g$ is a polynomial and $g(0)=0$ . $0$ is root of $g$ and hence $g(x)=x h(x)$ for some polynomial $h(x)$ . It will be enough if one can prove that $h(x)$ is constant . I'm stuck here .   Any form help will be highly appreciated.
98869
How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?  I have already recognized one problem caused from this: it  follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?  Best regards and thanks, John
105572
Let's assume the functions in question are continuously differentiable. If $f(g(x)) = f(x)g(x)$, then: $$f'(g(x))g'(x) = f'(x)g(x)+f(x)g'(x)$$ I attempted to find a particular solution by letting $g(x) = x$. Then, $$f'(x) = xf'(x) + f(x)$$ $$df = xdf + fdx$$ $$\frac{df}{f} = \frac{dx}{1-x}$$ $$\ln(f(x)) = -\ln(1-x)+C$$ $$f(x) = \frac {A}{1-x}$$ Where $A$ is a constant. This solution was fairly unsatisfying to me, and upon checking it, I see it doesn't actually work, as letting $g(x) = x$ leads to $f(x) = xf(x)$. I was wondering if there were better answers, or at least a less trivial way to go about the problem.
98334
Give $f (x) = x^3 - 4x^2 - 3x + 18$  For which values of $x$, will $f (x) . f'(x) &lt; 0$?  And also, how can I understand these type of questions? I always get stuck on them. Other examples are:  $$\frac{f(x)}{f'(x)} &gt; 0$$ $$xf (x) &gt; 0$$
31366
Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$:   $$f(f(x)f(y))+f(x+y)=f(xy).$$   It was six hours ago in IMO 2017 (problem 2).  I tried  the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.   Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?   Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and  $1-(1-x)(1-y)+1-x-y=1-xy$.
24854
How can I solve this functional equation, where $x,y$ are any real numbers and $f:\mathbb{R}\to \mathbb R$ is a function such that : $$f(x+y)+f(x-y)=2f(x)\cos y$$    I tried substituting $x=0$ to get $f(y)+f(-y)=2f(0)\cos y$ . Taking $x=y$ gives $f(2x)+f(0)=2f(x)\cos x$. I similar get some more relations like that, but its not really helping me in finding anything useful. I also think the function must either be $\sin$ or $\cos$ looking at the product to sum formulae. Can anyone tell me how should I solve this?
97943
Let $f(x)$ be a function such that $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$. Then for what value of $k$ is the relation $f(x + k) = f(x)$ necessarily true for every real $x$?  The answer/solution suggested in my module is like this: "this is a bit involved but can be proved that $k=6$". Could anybody explain me this?
14402
Let $x=0\,\, f(x+y)=f(0+y)$ every $y \in \mathbb{Z}$ So we get $f(0)+f(y)=f(y)$ then $f(0)=0$  $0=f(0)=f(x(-x))=f(x)+f(-x)=0$ Let $f(1)=k$ for some $k \in \mathbb{Z}$ $f(n)=f(1+1+1+...+1) (n \text{ many } 1)$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=f(1)+f(1)+...+f(1) (n \text{ many } f(1))$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=n\cdot k$  Thus $f: \mathbb{Z} \to \mathbb{Z}$ a function such that $f(n)=nk$ I found this but I must also show that there is no other function which satisfies $f(x+y)=f(x)+f(y)$ and I am stuck can somebody help me.