1. **Problem:** Find the Domain and Range of the functions: $$h(x) = \frac{\sqrt{4 - x^2}}{x - 3}$$ $$f(x) = \frac{1}{x} + \frac{5}{x - 3}$$ 2. **Domain and Range Concepts:** - The **domain** is the set of all $x$ values for which the function is defined. - The **range** is the set of all possible output values $y$. - For square roots, the expression inside must be $\geq 0$. - For rational functions, denominators cannot be zero. 3. **Find Domain of $h(x)$:** - Inside the square root: $4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2$ - Denominator: $x - 3 \neq 0 \implies x \neq 3$ - Since $3$ is outside $[-2,2]$, domain is $[-2,2]$ 4. **Find Range of $h(x)$:** - Numerator $\sqrt{4 - x^2}$ ranges from $0$ to $2$ (max at $x=0$). - Denominator $x-3$ ranges from $-5$ to $-1$ on $[-2,2]$ (always negative). - So $h(x)$ is negative or zero. - At $x=0$, $h(0) = \frac{2}{-3} = -\frac{2}{3}$ (max value). - At $x=\pm 2$, numerator $=0$, so $h(\pm 2) = 0$. - Range is $[-\frac{2}{3}, 0]$ 5. **Find Domain of $f(x)$:** - Denominators: $x \neq 0$, $x - 3 \neq 0 \implies x \neq 0, 3$ - Domain: $(-\infty, 0) \cup (0, 3) \cup (3, \infty)$ 6. **Range of $f(x)$:** - $f(x)$ is sum of two rational functions. - As $x \to 0^+$, $1/x \to +\infty$, $5/(x-3) \to -5/3$ finite, so $f(x) \to +\infty$. - As $x \to 0^-$, $1/x \to -\infty$, so $f(x) \to -\infty$. - As $x \to 3^+$, $5/(x-3) \to +\infty$, so $f(x) \to +\infty$. - As $x \to 3^-$, $5/(x-3) \to -\infty$, so $f(x) \to -\infty$. - As $x \to \pm \infty$, $f(x) \to 0$. - Range is all real numbers $(-\infty, \infty)$. **Final answers:** - Domain of $h(x)$: $[-2, 2]$ - Range of $h(x)$: $[-\frac{2}{3}, 0]$ - Domain of $f(x)$: $(-\infty, 0) \cup (0, 3) \cup (3, \infty)$ - Range of $f(x)$: $(-\infty, \infty)$