Consider f : R+ → [−5, ∞) given by f (x) = 9x2 + 6x − 5. Show

1.	Consider f : R+ → [−5, ∞) given by f (x) = 9x2 + 6x − 5. Show that f is invertible with f-1(y) = (√(y + 6) - 1)/3.
Answer» Here, function f : R₊ →[−5,∞) is given as f(x) = 9x²+ 6x − 5. Let y be any arbitrary element of [−5,∞). Let y = 9x² + 6x − 5 ⇒ y = (3x + 1)² − 1 − 5 = (3x + 1)² − 6 ⇒ (3x + 1)² = y + 6 ⇒ (3x + 1)² = √(y + 6) [as y ≥ − 5 ⇒ y + 6 ≥ 0] x = (√(y = 6) - 1)/3 Therefore, f is onto, thereby range f = [−5,∞). Let us define g : [−5,∞)→ R+ as g(y) = (√(y = 6) - 1)/3 Now, (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 Therefore,gof = I_R+ and fog = I_[−5,∞) Hence, f is invertible and the inverse of f if given by f^-1(y) = g(y) = (√(y + 6) - 1)/3

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

Answer»

Here, function f : R₊ →[−5,∞) is given as f(x) = 9x²+ 6x − 5.

Let y be any arbitrary element of [−5,∞).

Let y = 9x² + 6x − 5

⇒ y = (3x + 1)² − 1 − 5 = (3x + 1)² − 6

⇒ (3x + 1)² = y + 6

⇒ (3x + 1)² = √(y + 6) [as y ≥ − 5 ⇒ y + 6 ≥ 0]

x = (√(y = 6) - 1)/3

Therefore, f is onto, thereby range f = [−5,∞).

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

Now, (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

Therefore,gof = I_R+ and fog = I_[−5,∞)

Hence, f is invertible and the inverse of f if given by f^-1(y) = g(y) = (√(y + 6) - 1)/3