text: # What are all even  functions in real set which are true in relation f(x+y)=(f(x)+f(y))/(1+f(x)*f(y))?
  
  embizze | Certified Educator
  
  Find all even functions `f:RR->RR` such that `f(x+y)=(f(x)+f(y))/(1+f(x)f(y))`
  
  There are three such functions:
  
  Let x=y=0. Then `f(0+0)=f(0)=(f(0)+f(0))/(1+f(0)f(0))` so
  
  `f(0)=(2f(0))/(1+f(0)^2)`
  
  `=>f(0)^3-f(0)=0 ==> f(0)=0,+-1`
  
  (1) Suppose f(0)=0. Let y=-x. Note that f(x)=f(-x) since the function is even.
  
  Then `f(x+y)=f(x-x)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`
  
  then `0=(2f(x))/(1+f(x)^2)` . The denominator is never zero so this is true when 2f(x)=0 ==>f(x)=0.
  
  The first function is `f(x)=0 AA x in RR`
  
  (2) Suppose f(0)=1. Let y=-x.
  
  Then `f(x+y)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`
  
  and `1=(2f(x))/(1+f(x)^2)` ==> `f(x)^2-2f(x)+1=0`
  
  `==> (f(x)-1)^2=0 ==> f(x)=1`
  
  The second function is `f(x)=1 AA x in RR`
  
  (3) Suppose f(0)=-1. Let y=-x.
  
  Then `f(x+y)=f(x-x)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`
  
  so `-1=(2f(x))/(1+f(x)^2) ==> f(x)^2+2f(x)+1=0`
  
  then `(f(x)+1)^2=0 ==> f(x)=-1`
  
  The third function is `f(x)=-1 AA x in RR`
explanation: The document contains a math question and a detailed answer. The question asks to find all even functions in the set of real numbers that satisfy the relation \( f(x+y) = \frac{f(x) + f(y)}{1 + f(x)f(y)} \). The answer provides a step-by-step solution, identifying three such functions: \( f(x) = 0 \), \( f(x) = 1 \), and \( f(x) = -1 \) for all \( x \) in the real numbers.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # What are all even  functions in real set which are true in relation f(x+y)=(f(x)+f(y))/(1+f(x)*f(y))?
  
  embizze | Certified Educator
  
  Find all even functions `f:RR->RR` such that `f(x+y)=(f(x)+f(y))/(1+f(x)f(y))`
  
  There are three such functions:
  
  Let x=y=0. Then `f(0+0)=f(0)=(f(0)+f(0))/(1+f(0)f(0))` so
  
  `f(0)=(2f(0))/(1+f(0)^2)`
  
  `=>f(0)^3-f(0)=0 ==> f(0)=0,+-1`
  
  (1) Suppose f(0)=0. Let y=-x. Note that f(x)=f(-x) since the function is even.
  
  Then `f(x+y)=f(x-x)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`
  
  then `0=(2f(x))/(1+f(x)^2)` . The denominator is never zero so this is true when 2f(x)=0 ==>f(x)=0.
  
  The first function is `f(x)=0 AA x in RR`
  
  (2) Suppose f(0)=1. Let y=-x.
  
  Then `f(x+y)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`
  
  and `1=(2f(x))/(1+f(x)^2)` ==> `f(x)^2-2f(x)+1=0`
  
  `==> (f(x)-1)^2=0 ==> f(x)=1`
  
  The second function is `f(x)=1 AA x in RR`
  
  (3) Suppose f(0)=-1. Let y=-x.
  
  Then `f(x+y)=f(x-x)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`
  
  so `-1=(2f(x))/(1+f(x)^2) ==> f(x)^2+2f(x)+1=0`
  
  then `(f(x)+1)^2=0 ==> f(x)=-1`
  
  The third function is `f(x)=-1 AA x in RR`
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer. The question asks to find all even functions in the set of real numbers that satisfy the relation \( f(x+y) = \frac{f(x) + f(y)}{1 + f(x)f(y)} \). The answer provides a step-by-step solution, identifying three such functions: \( f(x) = 0 \), \( f(x) = 1 \), and \( f(x) = -1 \) for all \( x \) in the real numbers.
  
  Label: question-and-answer