Given function f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x – 5
Let us show that f is invertible.
Injectivity of f:
Let x and y be any two elements of domain (R+),
Such that f(x) = f(y)
⇒ 9x2 + 6x – 5 = 9y2 + 6y − 5
⇒ 9x2 + 6x = 9y2 + 6y
⇒ x = y (As, x, y ∈ R+)
Therefore, f is one-one.
Surjectivity of f:
Let y is in the co domain (Q)
Such that f(x) = y
⇒ 9x2 + 6x – 5 = y
⇒ 9x2 + 6x = y + 5
⇒ 9x2 + 6x + 1 = y + 6 (adding 1 on both sides)
⇒ (3x + 1)2 = y + 6
⇒ 3x + 1 =√(y + 6)
⇒ 3x = √(y + 6) – 1
⇒ x = (√(y + 6) -1)/3 in R+ (domain)
f is onto.
Therefore, f is a bijection and hence, it is invertible.
Let us find f-1
Let f−1(x) = y ...(1)
⇒ x = f(y)
⇒ x = 9y2 + 6y − 5
⇒ x + 5 = 9y2 + 6y
⇒ x + 6 = 9y2 + 6y + 1 (By adding 1 on both sides)
⇒ x + 6 = (3y + 1)2
⇒ 3y + 1 = √(x + 6)
⇒ 3y =√(x + 6) - 1
⇒ y = (√(x + 6) - 1)/3
Put these values in 1 we get,
Therefore, f-1(x) = (√(x - 6) - 1)/3