Question -
Answer -
Given f: R+ →[−5, ∞) given by f(x) = 9x2 + 6x – 5
We have to show that fis invertible.
Injectivity of f:
Let x and y be 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+)
So, f isone-one.
Surjectivityof 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 (By adding 1 on both sides)
⇒ (3x + 1)2 =y + 6
⇒ 3x + 1 = √(y + 6)
⇒ 3x = √ (y + 6) – 1
⇒ x = (√ (y + 6)-1)/3in R+ (domain)
f is onto.
So, f is a bijection and hence, it is invertible.
Now we have to 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 (adding 1 on both sides)
⇒ x + 6 = (3y + 1)2
⇒ 3y + 1 = √ (x + 6)
⇒ 3y =√(x +6) -1
⇒ y = (√ (x+6)-1)/3
Now substituting this valuein (1) we get,
So, f-1(x) =(√ (x+6)-1)/3