1. Find the domain and range of $f(x) = 1 + x^2$. - Domain: Since $x^2$ is defined for all real $x$, domain is all real numbers: $(-\infty, \infty)$. - Range: $x^2 \geq 0$, so $1 + x^2 \geq 1$. Range is $[1, \infty)$. 2. Find the domain and range of $f(x) = 1 - \sqrt{x}$. - Domain: The expression under the square root must be $\geq 0$, so $x \geq 0$. - Range: $\sqrt{x} \geq 0$, so $1 - \sqrt{x} \leq 1$. Minimum value when $x=0$ is $1$, and as $x \to \infty$, $f(x) \to -\infty$. Range is $(-\infty, 1]$. 3. Find the domain and range of $F(x) = \sqrt{5x + 10}$. - Domain: $5x + 10 \geq 0 \Rightarrow x \geq -2$. - Range: Square root outputs $\geq 0$, so range is $[0, \infty)$. 4. Find the domain and range of $g(x) = \sqrt{x^2 - 3x}$. - Domain: Inside the root $x^2 - 3x \geq 0$. Factor: $x(x - 3) \geq 0$. This holds when $x \leq 0$ or $x \geq 3$. - Range: Since square root is $\geq 0$, minimum is 0. At $x=0$ or $x=3$, $g(x)=0$. As $x \to \pm \infty$, $g(x) \to \infty$. Range is $[0, \infty)$. 5. Find the domain and range of $f(t) = \frac{4}{3 - t}$. - Domain: Denominator $3 - t \neq 0 \Rightarrow t \neq 3$. - Range: $f(t)$ can take all real values except where denominator zero. As $t \to 3$, $f(t) \to \pm \infty$. Range is $(-\infty, \infty)$. 6. Find the domain and range of $G(t) = \frac{2}{t^2 - 16}$. - Domain: Denominator $t^2 - 16 \neq 0 \Rightarrow t \neq \pm 4$. - Range: The function can take all real values except 0 because numerator is 2. As $t \to \pm 4$, $G(t) \to \pm \infty$. Range is $(-\infty, 0) \cup (0, \infty)$. Final answers: 1. Domain: $(-\infty, \infty)$, Range: $[1, \infty)$ 2. Domain: $[0, \infty)$, Range: $(-\infty, 1]$ 3. Domain: $[-2, \infty)$, Range: $[0, \infty)$ 4. Domain: $(-\infty, 0] \cup [3, \infty)$, Range: $[0, \infty)$ 5. Domain: $(-\infty, 3) \cup (3, \infty)$, Range: $(-\infty, \infty)$ 6. Domain: $(-\infty, -4) \cup (-4, 4) \cup (4, \infty)$, Range: $(-\infty, 0) \cup (0, \infty)$