query: Is identity the only function $f$ on real line satisfying $ f(x +f(y) + yf(x)) = y +f(x) + xf(y) ,\forall x,y \in \mathbb R$?
positive_passages.docid: 27809
positive_passages.text: Find the function equation $f(x+f(y)+yf(x))=y+f(x)+xf(y)$
negative_passages.docid: 132113
negative_passages.text: Does $f(x) = f(2x)$ for all real $x$, imply that $f(x)$ is a constant function?
negative_passages.docid: 62966
negative_passages.text: Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$
negative_passages.docid: 69734
negative_passages.text: how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$
negative_passages.docid: 129599
negative_passages.text: Functional Equation: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $(x+y)(f(x)-f(y))=(x-y)f(x+y)$
negative_passages.docid: 50111
negative_passages.text: non-continuous function satisfies $f(x+y)=f(x)+f(y)$
negative_passages.docid: 14495
negative_passages.text: I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$
negative_passages.docid: 149328
negative_passages.text: A function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$ satisfies $f(z)+f(-z)=0$.(MADHAVA-2020)
negative_passages.docid: 103934
negative_passages.text: Functional Equation (no. of solutions): $f(x+y) + f(x-y) = 2f(x) + 2f(y)$
negative_passages.docid: 59063
negative_passages.text: $f(f(x))=f(x)$ question
negative_passages.docid: 11828
negative_passages.text: continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$
negative_passages.docid: 125788
negative_passages.text: A function with a property $f(x+y)=f(x)f(y)$
negative_passages.docid: 48925
negative_passages.text: BMO1 2009/10 Q5 functional equation: $f(x)f(y) = f(x + y) + xy$
negative_passages.docid: 32856
negative_passages.text: Functional equations $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$
negative_passages.docid: 133682
negative_passages.text: Prove that if $f(t)$ satisfies $f(t)=f'(t)$, then $f(t+s)=f(t)f(s)$ for all $t,s\in R$ without using the fact that $f(t)=c\cdot e^t$.
negative_passages.docid: 46875
negative_passages.text: The identity morphism in $\mathbf{Set}$ is the identity function
negative_passages.docid: 121729
negative_passages.text: Prove without using graphing calculators that $f: \mathbb R\to \mathbb R,\,f(x)=x+\sin x$ is both one-to-one, onto (bijective) function.