1. **Problem:** Find the domain and range of the rational function $g(x) = \frac{2}{x - 5}$. 2. **Domain:** The domain is all real numbers except where the denominator is zero. Set denominator equal to zero: $$x - 5 = 0 \implies x = 5$$ So, domain is: $$\{x \mid x \neq 5\}$$ 3. **Range:** For $g(x) = \frac{2}{x - 5}$, the function can take any real value except zero because the numerator is constant and denominator can be any real number except zero. Set $y = \frac{2}{x - 5}$ and solve for $x$: $$y = \frac{2}{x - 5} \implies x - 5 = \frac{2}{y} \implies x = 5 + \frac{2}{y}$$ For $x$ to be real, $y \neq 0$. So, range is: $$\{y \mid y \neq 0\}$$ --- 1. **Problem:** Find the domain and range of the rational function $h(x) = \frac{2x}{(x + 2)(x - 2)}$. 2. **Domain:** Denominator cannot be zero: $$(x + 2)(x - 2) = 0 \implies x = -2 \text{ or } x = 2$$ Domain: $$\{x \mid x \neq -2, x \neq 2\}$$ 3. **Range:** Set $y = \frac{2x}{(x + 2)(x - 2)} = \frac{2x}{x^2 - 4}$. Rewrite: $$y = \frac{2x}{x^2 - 4} \implies y(x^2 - 4) = 2x \implies yx^2 - 4y = 2x$$ Rearranged: $$yx^2 - 2x - 4y = 0$$ Consider as quadratic in $x$: $$yx^2 - 2x - 4y = 0$$ Discriminant: $$\Delta = (-2)^2 - 4(y)(-4y) = 4 + 16y^2 = 4(1 + 4y^2) > 0$$ Since $\Delta > 0$ for all real $y$, there are real $x$ solutions for all $y$. Check if $y=0$ is possible: $$0 = \frac{2x}{x^2 - 4} \implies 2x = 0 \implies x=0$$ $x=0$ is in domain, so $y=0$ is in range. Therefore, range is all real numbers: $$\{y \mid y \in \mathbb{R}\}$$ --- 1. **Problem:** Find the domain and range of the rational function $r(x) = \frac{x^2 - 4}{-x^2 - 2x + 3}$. 2. **Domain:** Denominator cannot be zero: $$-x^2 - 2x + 3 = 0 \implies x^2 + 2x - 3 = 0$$ Solve quadratic: $$x = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm 4}{2}$$ Roots: $$x = 1 \text{ or } x = -3$$ Domain: $$\{x \mid x \neq 1, x \neq -3\}$$ 3. **Range:** Set $y = \frac{x^2 - 4}{-x^2 - 2x + 3}$. Rewrite denominator: $$-x^2 - 2x + 3 = -(x^2 + 2x - 3)$$ So, $$y = \frac{x^2 - 4}{-(x^2 + 2x - 3)} = -\frac{x^2 - 4}{x^2 + 2x - 3}$$ Cross-multiplied: $$y(x^2 + 2x - 3) = -(x^2 - 4)$$ Expand: $$yx^2 + 2yx - 3y = -x^2 + 4$$ Bring all terms to one side: $$(y + 1)x^2 + 2yx - 3y - 4 = 0$$ Consider as quadratic in $x$: $$A = y + 1, B = 2y, C = -3y - 4$$ Discriminant: $$\Delta = B^2 - 4AC = (2y)^2 - 4(y + 1)(-3y - 4) = 4y^2 + 4(y + 1)(3y + 4)$$ Calculate: $$4y^2 + 4[(y)(3y + 4) + 1(3y + 4)] = 4y^2 + 4(3y^2 + 4y + 3y + 4) = 4y^2 + 4(3y^2 + 7y + 4)$$ $$= 4y^2 + 12y^2 + 28y + 16 = 16y^2 + 28y + 16$$ For real $x$, $\Delta \geq 0$: $$16y^2 + 28y + 16 \geq 0$$ Divide by 4: $$4y^2 + 7y + 4 \geq 0$$ Find roots: $$y = \frac{-7 \pm \sqrt{49 - 64}}{8} = \frac{-7 \pm \sqrt{-15}}{8}$$ Discriminant negative, so quadratic always positive. Therefore, $\Delta > 0$ for all real $y$. Check if denominator $A = y + 1 = 0$ (to avoid division by zero in quadratic formula): If $y = -1$, then equation becomes: $$0 \cdot x^2 + 2(-1)x - 3(-1) - 4 = -2x + 3 - 4 = -2x - 1 = 0$$ Solve: $$-2x - 1 = 0 \implies x = -\frac{1}{2}$$ Check if $x = -\frac{1}{2}$ is in domain: Denominator at $x = -\frac{1}{2}$: $$-\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 3 = -\frac{1}{4} + 1 + 3 = 3.75 \neq 0$$ So $y = -1$ is in range. Hence, range is all real numbers: $$\{y \mid y \in \mathbb{R}\}$$ --- 1. **Problem:** Find the domain and range of the rational function $m(x) = \frac{x^2 - x - 12}{x^2 - 3x - 4}$. 2. **Domain:** Denominator cannot be zero: $$x^2 - 3x - 4 = 0$$ Factor: $$(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1$$ Domain: $$\{x \mid x \neq 4, x \neq -1\}$$ 3. **Range:** Set $$y = \frac{x^2 - x - 12}{x^2 - 3x - 4}$$ Factor numerator and denominator: $$x^2 - x - 12 = (x - 4)(x + 3)$$ $$x^2 - 3x - 4 = (x - 4)(x + 1)$$ Simplify: $$y = \frac{(x - 4)(x + 3)}{(x - 4)(x + 1)}$$ For $x \neq 4$, cancel $(x - 4)$: $$y = \frac{x + 3}{x + 1}$$ Rewrite: $$y = \frac{x + 3}{x + 1} \implies y(x + 1) = x + 3 \implies yx + y = x + 3$$ Rearranged: $$(y - 1)x = 3 - y$$ If $y \neq 1$: $$x = \frac{3 - y}{y - 1}$$ Check if $x$ is in domain: $x \neq 4$ and $x \neq -1$. Set $x = 4$: $$4 = \frac{3 - y}{y - 1} \implies 4(y - 1) = 3 - y \implies 4y - 4 = 3 - y \implies 5y = 7 \implies y = \frac{7}{5}$$ Set $x = -1$: $$-1 = \frac{3 - y}{y - 1} \implies -1(y - 1) = 3 - y \implies -y + 1 = 3 - y \implies 1 = 3$$ Contradiction, so $y = \frac{7}{5}$ is the only value excluded from range. If $y = 1$: $$y = 1 = \frac{x + 3}{x + 1} \implies x + 3 = x + 1 \implies 3 = 1$$ Contradiction, so $y = 1$ is not in range. Therefore, range excludes $y = 1$ and $y = \frac{7}{5}$: $$\{y \mid y \neq 1, y \neq \frac{7}{5}\}$$ --- **Final answers in set builder notation:** - $g(x)$ domain: $\{x \mid x \neq 5\}$ - $g(x)$ range: $\{y \mid y \neq 0\}$ - $h(x)$ domain: $\{x \mid x \neq -2, x \neq 2\}$ - $h(x)$ range: $\{y \mid y \in \mathbb{R}\}$ - $r(x)$ domain: $\{x \mid x \neq 1, x \neq -3\}$ - $r(x)$ range: $\{y \mid y \in \mathbb{R}\}$ - $m(x)$ domain: $\{x \mid x \neq 4, x \neq -1\}$ - $m(x)$ range: $\{y \mid y \neq 1, y \neq \frac{7}{5}\}$