Here Are The Solution Step-by-step:

1. Identify the given functions and their domains/ranges We are given two functions: $$ f(x) = x^2 - 1 $$ $$ g(x) = \sqrt{x} + 2 $$ * Domain of $f$: Since $f(x) = x^2 - 1$ is a standard polynomial, its domain is all real numbers: $$\text{Domain of } f = (-\infty, \infty)$$ * Range of $f$: The minimum value of $x^2 - 1$ occurs at $x = 0$, giving $f(0) = -1$. Thus, $f(x) \ge -1$: $$\text{Range of } f = [-1, \infty)$$ * Domain of $g$: Since $g(x)$ contains a square root, we require $x \ge 0$: $$\text{Domain of } g = [0, \infty)$$ * Range of $g$: Since $\sqrt{x} \ge 0$, we have $\sqrt{x} + 2 \ge 2$: $$\text{Range of } g = [2, \infty)$$ --- ### 2. Determine $(f \circ g)(x)$ and its Domain The composite function $(f \circ g)(x)$ is defined as $f(g(x))$. #### Formula: $$ (f \circ g)(x) = f(g(x)) = f(\sqrt{x} + 2) $$ Substitute $g(x) = \sqrt{x} + 2$ into $f(x) = x^2 - 1$: $$ (f \circ g)(x) = (\sqrt{x} + 2)^2 - 1 $$ Expand the squared term: $$ (f \circ g)(x) = (x + 4\sqrt{x} + 4) - 1 $$ $$ (f \circ g)(x) = x + 4\sqrt{x} + 3 $$ #### Domain of $(f \circ g)(x)$: The domain of $(f \circ g)$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$. 1. $x$ must be in the domain of $g$, so $x \ge 0$. 2. $g(x) = \sqrt{x} + 2$ must be in the domain of $f$, which is $(-\infty, \infty)$. This is satisfied for all $x \ge 0$. Therefore, the domain of $(f \circ g)$ is: $$ \text{Domain of } (f \circ g) = [0, \infty) $$ --- ### 3. Determine $(g \circ f)(x)$ and its Domain The composite function $(g \circ f)(x)$ is defined as $g(f(x))$. #### Formula: $$ (g \circ f)(x) = g(f(x)) = g(x^2 - 1) $$ Substitute $f(x) = x^2 - 1$ into $g(x) = \sqrt{x} + 2$: $$ (g \circ f)(x) = \sqrt{x^2 - 1} + 2 $$ #### Domain of $(g \circ f)(x)$: The domain of $(g \circ f)$ consists of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$. 1. $x$ must be in the domain of $f$, which is $(-\infty, \infty)$. 2. $f(x)$ must be in the domain of $g$, which means $f(x) \ge 0$: $$ x^2 - 1 \ge 0 $$ $$ (x - 1)(x + 1) \ge 0 $$ This inequality holds when $x \le -1$ or $x \ge 1$. Therefore, the domain of $(g \circ f)$ is: $$ \text{Domain of } (g \circ f) = (-\infty, -1] \cup [1, \infty) $$ --- ### Final Answer: 1. $(f \circ g)(x) = x + 4\sqrt{x} + 3$ with Domain: $[0, \infty)$ 2. $(g \circ f)(x) = \sqrt{x^2 - 1} + 2$ with Domain: $(-\infty, -1] \cup [1, \infty)$