(iii) $g(x) = \begin{cases} 6x + 7, & x \le -2 \\ 4 - 3x, & x > -2 \end{cases}$

(iii) $g(x) = \begin{cases} 6x + 7, & x \le -2 \\ 4 - 3x, & x > -2 \end{cases}$

Given:

• Domain ($D_g$): Both intervals combine to cover all real numbers:

  • For $x \le -2$: $g(-2) = 6(-2) + 7 = -5$, $g(-3) = 6(-3) + 7 = -11$
  • For $x > -2$: $g(0) = 4 - 3(0) = 4$, $g(2) = 4 - 3(2) = -2$
  $$D_g = (-\infty, -2] \cup (-2, \infty) = \mathbb{R}$$ $$\boxed{D_g = \mathbb{R} = (-\infty, \infty)}$$

• Range ($R_g$):

  • For $x \le -2$: $g(x) \le 6(-2) + 7 = -5 \implies (-\infty, -5]$
  • For $x > -2$: $g(x) < 4 - 3(-2) = 10 \implies (-\infty, 10)$ Combined range:
  $$R_g = (-\infty, -5] \cup (-\infty, 10) = (-\infty, 10)$$ $$\boxed{R_g = (-\infty, 10)}$$