Find all real functions $f\colon \mathbb{R}\rightarrow \mathbb{R}$ so that $(x+y)(f(x)-f(y))=f(x^2)-f(y^2)$.  Can someone at least find the value of $f(1)$ if it is possible, it would help me.
62941
Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ such that $$(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$$  I've tried subbing in heaps of values but I keep getting things like $f(0)=f(0)$ and other such useless results.  Any help would be hugely appreciated.
77195
Does there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y \in\mathbb{R}$; $f(1)=1$; and $f(\sqrt 2)=3$?  Intuitively, I believe the answer is no. However, I'm having a hard time proving this. My initial thought was a proof by contradiction. Somehow using  $f(1)=1$ and $f(\sqrt 2)=3$ to show that $f(x+y)\neq(x)+f(y)$ for some choice in $x$ and $y$. This didn't prove fruitful. My next thought was to use properties of functions satisfying the addition property of homomorphisms to prove that $f(\sqrt 2)\neq3$. For example, one such property is $f(nx)=nf(x)$ for all $n \in \mathbb{Z}, x \in \mathbb{R}$. With this, one could theoretically construct a proof by induction that $f(x)=x$ for all $x \in \mathbb{R}$. Once again, this didn't prove fruitful. After these failed attempts at showing such a function doesn't exist, I considered how to show that one does exist. For reference, this question was asked in the context of a Field Theory course. As such, the thought for how to construct a proof is to satisfy the addition property with some linear transformation and show that a correct selection of basis spans all of $\mathbb{R}$. This idea seems feasible, however I am not sure where to begin. Any advice on how to approach this problem is greatly appreciated.
79524
Find all continuous functions such that $f(x+y) = f(x)f(y)f(xy)$   I solved it and got the two very obvious solutions, $f(x) = 0$ and $f(x) = 1$. Any other such functions?
41993
Let $f:\mathbb{R}\to \mathbb{R}$ is a function such that for all real $x$ and $y$, $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$, then prove that $f$ must be one of the two following functions:   $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=0$ for all real $x$   OR   $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ for all real $x$   I got to the point where putting the two equations together, you get $f(x+y)f(x)= f(xy) + f(x)^2$ and plugging in $f(x)=x$ checks with it. So am I going in the right direction or am I just doing some guess work? Is there a more elegant way of doing it?   Thanks
114828
Let $f: \mathbb{R} \to D \subseteq \mathbb{R}$ be a continuous function. Is there a function $f$ that satisfies the following property? $\forall y \in D$, there are exactly 3 $ x_1,x_2,x_3 \in \mathbb{R}$ such that $f(x_1) = f(x_2) = f(x_3) = y$.
81155
Suppose a function $f : \mathbb{R} \to \mathbb{R} $ satisfies the relation $$f(x^2) = xf(x) \ \ \forall x$$  Does this imply $f$ must be a straight line, $f(x) = mx$? If so, why?  If not, are there other such functions?
23743
Let $f,g:$ $R \rightarrow R$ where $g(x)$ $=$ $1$ $-$ $x$ $+$ $x^2$ and $f(x)$ $=$ $ax$ $+$ $b$  If $g(f(x))$ $=$ $9x^2$ $-$ $9x$ $+$ $3$, determine $a$ and $b$.  So far, I have "fit" $f(x)$ into $g(x)$ as follows: $g(f(x))$ $=$ $(ax + b)^2$ $-$ $(ax+b)$ $+$ $1$  Is there a relationship between compositions and inverse functions I should be using to isolate $a$ and $b$? How does one extract those variables in order to determine their value?  Thank you!
43176
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be such that    $$f\big(x-f(y)\big)=f\big(f(y)\big)+x\cdot f(y)+f(x)-1$$   for all $x,y \in\mathbb{R}$.  Then, what is the value of $|f(16)|-25$?   what i try  put $f(y)=z.$ then $f(x-z)=f(f(x))+xz+f(x)-1$  put $f(x)=t.$ then $f(x-z)=f(t)+zx+t-1$  How do i solve it Help me please
72804
I am wondering what is the class of functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that $f(f(x))=f(x)$?  I think it should be:   Constant Value functions the identity function absolute value function $|x|$   But I don't know if this is right or how to show it rigorously.  Any suggestions?
51157
Find all functions $f$ such that $f:\mathbb{N}\rightarrow\  \mathbb{N}$ and $f(f(n))+f(n+1)=n+2$   Let us plug in $n=1$  $f(f(1))+f(2)=3$  Since the function is from $\mathbb{N}$ to $\mathbb{N}$, $f(2)$ can only take the values $1,2$. Now we divide the problem into cases.  Case-1: $f(f(1))=2,f(2)=1$  We can assume that $f(1)=c$ for the time being. Then plugging in $n=3$ and using $f(2)=1$ gives $$f(3)=4-c$$ and again since the range of the function is positive integers,then $4-c$ has to be positive and hence $c$ belongs to  {$1,2,3$}. Now, $$f(1)=c$$$$\implies f(f(1))=f(c)$$$$\implies 2=f(c)$$ by the assumption that $f(f(1))=2$ . Now,since $c$ can only take the values $1,2,3$,we start treating cases. If $c=1$,$$f(c)=2$$$$\implies f(1)=2$$ but we know from the deinition of $c$ that $f(1)=c=1$,a contradiction.If $c=2$,then $2=f(c)=f(2)$ but $f(2)=1$ by assumption. Finally,if $c=3$ $$2=f(c)=f(3)$$ but $$f(3)=4-c=4-3=1$$ which is once again a contradiction. Therefore there are no such functions in this case.  Case-2: $f(f(1))=1,f(2)=2$  Again assuming $f(1)=c$ and using $f(n)\le n$ along with plugging $n=c-1$ will give us that $f(1)=1$ and then it is easy to prove that such a function exists by recursion. I can only give a "sort of recursive" way to define the function. Here it goes $$f:\mathbb{N}\rightarrow \mathbb{N}$$$$f(1)=1$$$$f(n)=n+1-f(f(n-1))$$  But this case is harder to deal with.Some help will be appreciated.
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!
47976
I must assume that there exists f such that:  $$f(x+y) = f(x)+f(y)\\f(x\cdot y) = f(x)\cdot f(y)$$ One thing I managed to do was:  $$f(2+2) = f(2) + f(2) = f(2\cdot 2) = f(2)\cdot f(2) \implies \\f(2)+f(2) = f(2)\cdot f(2)\implies \\2f(2) = f^2(2) \implies f^2(2)-2f(2)=0 \implies f(2)(f(2)-2) = 0 \implies\\ f(2) = 0 \text{ or } f(2) = 2$$  but how to extend this argument to all the integers? That is, how to force $f(3) = 3$ and so on?
114172
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.  What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?   I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?
3532
Does there exist a continuous function $\: f : \mathbf{R} \to \mathbf{R} \:$ such that for   all  functions $\: g : \mathbf{R} \to \mathbf{R} \:$, for all real numbers $x$,  there exists a real number $y$ such that $\: x &lt; y \:$ and $\: g(y) &lt; f(y) \:$?
19092
Find all functions $f$, defined on the real numbers and taking real values, which satisfy the equation $f(x)f(y) = f(x + y) + xy$ for all real numbers $x$ and $y$.  I worked out $f(0)=1$, and $f(-1)f(1)=0$, but then I hit a wall.
103853
The question asks us to find the function $f(x)$ with the given information     Let $f:\mathbb R   \rightarrow  \mathbb R$ such that $f'(0)=1$ and $f(x+y)=f(x) + f(y) + (e^{x+y})(x+y)-xe^x-ye^y+2xy$ where $x,y ∈\mathbb R$. Then determine $f(x)$.   Also,It will be great if someone could explain the general approach used in such questions.  Thanks in advance.
44696
Find functions $f,g : \mathbb R \longrightarrow \mathbb R $ and $a,b,c \in \mathbb R $ such that  $$ \lim_{x \rightarrow a} f(x) = b \quad \text{and} \quad \lim_{y \rightarrow b} g(y) = c \quad \text{but}\quad \lim_{x \rightarrow a} g(f(x)) \neq c$$  I tried to use non continous functions but i didnt manage to get a any useful results.