Given function f: R → R given by f(x) = 4x + 3
Let us show that the given function is invertible.
Consider injection of f:
Let x and y be any two elements of domain (R),
Such that f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
Therefore, f is one-one.
Now surjection of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ 4x + 3 = y
⇒ 4x = y - 3
⇒ x = (y - 3)/4 in R (domain)
⇒ f is onto.
Therefore, f is a bijection and, hence, is invertible.
Let us find f -1
Let f-1(x) = y ...(1)
⇒ x = f(y)
⇒ x = 4y + 3
⇒ x − 3 = 4y
⇒ y = (x - 3)/4
Now put these values in 1 we get
Therefore, f-1(x) = (x - 3)/4
