(v) $g(x) = \dfrac{x + 2}{3 - x}$

(v) $g(x) = \dfrac{x + 2}{3 - x}$

Given:

• Domain ($D_g$): Denominator cannot equal zero ($3 - x \ne 0 \implies x \ne 3$):

  • $g(-2) = \dfrac{-2 + 2}{3 - (-2)} = \dfrac{0}{5} = 0$
  • $g(0) = \dfrac{0 + 2}{3 - 0} = \dfrac{2}{3}$
  • For $x = 3$: $\dfrac{3 + 2}{3 - 3} = \dfrac{5}{0}$ (Undefined)
  $$\boxed{D_g = \mathbb{R} \setminus \{3\} = (-\infty, 3) \cup (3, \infty)}$$

• Range ($R_g$): Let $y = \dfrac{x + 2}{3 - x}$ and solve for $x$:

  $$\begin{aligned} y(3 - x) &= x + 2 \\ 3y - xy &= x + 2 \\ 3y - 2 &= x(1 + y) \\ x &= \frac{3y - 2}{1 + y} \end{aligned}$$

  For $x$ to be defined, require $1 + y \ne 0 \implies y \ne -1$:

  • For $y = 2 \implies x = \dfrac{3(2) - 2}{1 + 2} = \dfrac{4}{3} \implies g(4/3) = 2$
  $$\boxed{R_g = \mathbb{R} \setminus \{-1\} = (-\infty, -1) \cup (-1, \infty)}$$