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Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f-1(x) = (√(x + 6) - 1)/3 

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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)

⇒ 9x+ 6x – 5 = 9y+ 6y − 5

⇒ 9x+ 6x = 9y+ 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 = 9y+ 6y − 5

⇒ x + 5 = 9y+ 6y

⇒ x + 6 = 9y+ 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 

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