Question

f : [2, ∞) → (-∞, 4], where f(x) = x(4 – x) then f⁻¹(x) is given by

• A. 2 −$\sqrt{4 - x}$
• B. 2 + $\sqrt{4 - x}$
• C. −2$\sqrt{4 - x}$
• D. −2 −$\sqrt{4 - x}$

Answer 1

Answer: (B)

2 + $\sqrt{4 - x}$

Answer 2

f(x) = x(4 - x)

⇒ f(f-¹(x)) = f^-1(x) – f^-1(x))² = x

⇒ (f⁻¹(x))² - 4f^−l(x) + x = 0

$\frac { 4 \pm \sqrt { 16 - 4 x } } { 2 } = 2 \pm \sqrt { 4 - x }$

But f⁻¹_: (-∞, 4] → [2, ∞)

⇒ f^-1 : (x) = $2 + \sqrt { 4 - x }$