1. **State the problem:** Find the domain and range of the composition of two functions $g \circ f$, where $f$ and $g$ have given domains and ranges. 2. **Recall the composition definition:** The composition $g \circ f$ means applying $f$ first, then $g$ to the result: $$ (g \circ f)(x) = g(f(x)) $$ 3. **Domain of $g \circ f$:** The domain of $g \circ f$ consists of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$. 4. **Given sets:** - Domain of $f$: $\{1, 2, 3, 6, 7, 8\}$ - Range of $f$: $\{0, 4, 7, 8\}$ - Domain of $g$: $\{0, 3, 6, 7, 8, 9\}$ - Range of $g$: $\{7, 8, 9\}$ 5. **Check which values in the range of $f$ are in the domain of $g$:** Range of $f$ is $\{0, 4, 7, 8\}$ Domain of $g$ is $\{0, 3, 6, 7, 8, 9\}$ Only $4$ is not in the domain of $g$, so any $x$ such that $f(x) = 4$ must be excluded from the domain of $g \circ f$. 6. **Identify $x$ values where $f(x) = 4$:** From the arrows (or given info), $f$ maps some $x$ to $4$. Since the problem does not specify exactly which $x$ map to $4$, but the domain of $f$ is $\{1, 2, 3, 6, 7, 8\}$ and range includes $4$, we assume at least one $x$ maps to $4$. 7. **Domain of $g \circ f$ is all $x$ in domain of $f$ except those mapping to $4$:** So domain of $g \circ f = \{1, 2, 3, 6, 7, 8\} \setminus \{x: f(x) = 4\}$. 8. **Range of $g \circ f$:** The range of $g \circ f$ is the set of all $g(f(x))$ for $x$ in the domain of $g \circ f$. Since $f(x)$ takes values in $\{0, 4, 7, 8\}$ and $g$ maps domain $\{0, 3, 6, 7, 8, 9\}$ to range $\{7, 8, 9\}$, the possible outputs of $g(f(x))$ are in $\{7, 8, 9\}$. 9. **Final answers in set notation:** (a) Domain of $g \circ f = \{1, 2, 3, 6, 7, 8\} \setminus \{x: f(x) = 4\}$ (b) Range of $g \circ f = \{7, 8, 9\}$