(ii) $g(x) = \sqrt{x + 2}$

(ii) $g(x) = \sqrt{x + 2}$

Given:

• Domain ($D_g$): For real outputs, require the radicand to be non-negative ($x + 2 \ge 0 \implies x \ge -2$):

  • $g(-2) = \sqrt{-2 + 2} = \sqrt{0} = 0$
  • $g(2) = \sqrt{2 + 2} = \sqrt{4} = 2$
  • $g(7) = \sqrt{7 + 2} = \sqrt{9} = 3$
  • For $x = -3$: $\sqrt{-3 + 2} = \sqrt{-1} \notin \mathbb{R}$ (Undefined)
  $$\boxed{D_g = [-2, \infty)}$$

• Range ($R_g$): The principal square root produces non-negative outputs:

  $$y = \sqrt{x + 2} \ge 0$$

  Squaring both sides: $y^2 = x + 2 \implies x = y^2 - 2 \ge -2$.

  • Minimum output is $0$ at $x = -2$.
  • For $y = 5 \implies x = 5^2 - 2 = 23 \implies g(23) = 5$.
  $$\boxed{R_g = [0, \infty)}$$