For all x and y such that −1 < x < 1 (otherwise the denominator f X (x) vanishes) and < < (otherwise the conditional probability degenerates to 0 or 1) One may also treat the conditional probability as a random variable, — a function of the random variable X , namely,Therefore f(f − 1(x)) = f(y) = x A function f has an inverse function f − 1, iff f is bijective Let f A → B, such that f(x) = y, with x ∈ A, y ∈ B Then its inverse is a function such that f − 1 maps from the codomain of f to the domain of f, this is f − 1 B → A So, ∀y ∈ B, f − 1(y) = x, with x ∈ ALet x and y be any two elements in the domain (N), such that f(x) = f(y) ⇒ `x^2 x 1 = y^2 y 1` ⇒ `(x^2 y^2 ) (x y ) = 0 ` ⇒ (x y) (x y ) (xy ) = 0 ⇒ ( x y) ( x y 1) = 0 ⇒ x y = 0 x y 1 can not be zero because x and y are natural numbers ⇒ x =y So, f is oneone Surjectivity when x = 1 `x^2 x 1
How To Find Inverse Function Of X 1 X Given That F X Is Bigger Or Equal To 2 Quora
