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

Solution:

f : R₊ → [- 5, ∞) given by f (x) = 9x² + 6x - 5

Let y be an arbitrary element of [- 5, ∞).

Let y = 9x² + 6x - 5

⇒ y = (3x + 1)² - 1 - 5

⇒ y = (3x + 1)² - 6

⇒ (3x + 1)² = y + 6

⇒ 3x + 1 = √y + 6 [as y ≥ - 5 ⇒ y + 6 > 0]

⇒ x = [(√y + 6) - 1]/3

∴ f is onto, there by range f = [- 5, ∞).

Let us define g : [- 5, ∞) → R₊ as g (y) = [(√y + 6) - 1]/3

We have,

(gof)(x) = g (f (x)) = g (9x² + 6x - 5)

= g ((3x + 1)² - 6)

= (√(3x + 1)² - 6 + 6 - 1)/3

= (3x + 1 - 1)/3 = x

And,

(fog)(y) = f (g(y)) = F [((√y + 6) - 1)/3]

= [3 ((√y + 6) - 1)/3 + 1]² - 6

= (√y + 6)² - 6 = y + 6 - 6 = y

∴ gof = I_R and fog = I_{[- 5, ∞)}

Hence, f is invertible and the inverse of f is given by

f ^{- 1} (y) = g (y) = [(√y + 6) - 1]/3

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise 1.3 Question 9

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

Summary:

Considering f : R₊ → [- 5, ∞) given by f (x) = 9x² + 6x - 5. Hence we have shown that  f is invertible with f ^-1 (y) = ((√y + 6) - 1)/3