Composition of Functions, Domain of Composition - College Algebra

Let $f(x) = \frac{2x + 1}{x - 3}$ and $g(x) = \sqrt{x + 4}$ . What is the domain of the composition $(f \circ g)(x)$ ?
Answer: $[-4, \infty) \setminus \{5\}$ For $(f \circ g)(x) = f(g(x))$ , we need both $g(x)$ to be defined and $f(g(x))$ to be defined. First, $g(x) = \sqrt{x + 4}$ requires $x + 4 \geq 0$ , so $x \geq -4$ . Next, $f(g(x)) = \frac{2\sqrt{x + 4} + 1}{\sqrt{x + 4} - 3}$ requires the denominator to be nonzero: $\sqrt{x + 4} - 3 \neq 0$ , which means $\sqrt{x + 4} \neq 3$ . Since $\sqrt{x + 4} = 3$ when $x + 4 = 9$ , we get $x = 5$ . Therefore, the domain is $[-4, \infty) \setminus \{5\}$ .
Given the piecewise function $t(x) = \begin{cases} x^2 & \text{if } x \leq 0 \\ 2x - 1 & \text{if } x > 0 \end{cases}$ and $s(x) = x - 3$ , what is the domain of $(t \circ s)(x)$ ?
Answer: All real numbers For $(t \circ s)(x) = t(s(x)) = t(x - 3)$ , we need to determine when this composition is defined. The function $s(x) = x - 3$ is defined for all real numbers. The piecewise function $t(x)$ is also defined for all real numbers: $t(x) = x^2$ when $x \leq 0$ and $t(x) = 2x - 1$ when $x > 0$ . For the composition, we evaluate $t$ at $s(x) = x - 3$ . When $s(x) = x - 3 \leq 0$ (i.e., when $x \leq 3$ ), we use $t(s(x)) = (s(x))^2 = (x - 3)^2$ . When $s(x) = x - 3 > 0$ (i.e., when $x > 3$ ), we use $t(s(x)) = 2(s(x)) - 1 = 2(x - 3) - 1 = 2x - 7$ . Both pieces are defined for their respective domains, so the overall composition is defined for all real numbers.