$f (x) = \frac{4 - x^{2}}{3}$

Domain $\{x \in ℝ, x \geq 0\}$ , Range $\{y \in ℝ, y \leq \frac{4}{3}\}$

Find the inverse

\begin{aligned} L e t y & = f (x) \\ y & = \frac{4 - x^{2}}{3} \\ - 3 y + 4 & = x^{2} \\ \sqrt{- 3 y + 4} & = x \end{aligned}

Therefore

$f^{- 1} (x) = \sqrt{- 3 x + 4}$ Domain $\{x \in ℝ, x \leq \frac{4}{3}\}$ , Range $\{y \in ℝ, y \geq 0\}$