The resulting function of the given relation is f(x) = x² - 1 / 2
The term function is referred as the mathematical process that uniquely relates the value of one variable to the value of one (or more) other variables.
Here we need to determine all functions f:R→R such that f(x−f(y))=f(f(y))+xf(y)+f(x)−1∀x,y∈R
While we have clearly looking into the given problem, we have given that
=>f(x−f(y))=f(f(y))+xf(y)+f(x)−1(1)
Now, we have to Put x=f(y)=0, then we get the result as
=> f(0)=f(0)+0+f(0)−1
Therefore, the value of the function f(0)=1(2)
Now, again we have to put
=> x=f(y)=λ -------------(1)
Then we have to rewrite the relation like the following,
=> f(0)=f(λ)+λ²+f(λ)−1
=>1 = 2f(λ) + λ² − 1 -------------(2)
When we rewrite the function as,
=> f(λ) = λ² - 1 / 2
Therefore, the unique function is
=> f(x) = x² - 1 / 2
To know more about function here.
https://brainly.com/question/28193995
#SPJ4