1. **Evaluate (f \circ g)(6)** using the table. - First find $g(6)$ from the table: $g(6) = 4$. - Then find $f(g(6)) = f(4)$: from the table, $f(4) = 5$. So, $(f \circ g)(6) = 5$. 2. **Evaluate (f \circ f)(5)** using the table. - First find $f(5)$: from the table, $f(5) = 3$. - Then find $f(f(5)) = f(3)$: from the table, $f(3) = 4$. So, $(f \circ f)(5) = 4$. 3. **Evaluate $f(g(x))$ for each $x$ in the second table.** - For $x = -2$: $g(-2) = 5$, then $f(5) = 2$ (from the second table, $f(5)$ is not given, so we cannot evaluate this value). Since $f(5)$ is not in the table, $f(g(-2))$ is undefined. - For $x = -1$: $g(-1) = -1$, then $f(-1)$ is not in the table, so undefined. - For $x = 0$: $g(0) = 3$, then $f(3) = 2$. - For $x = 3$: $g(3) = 0$, then $f(0) = -1$. - For $x = 4$: $g(4) = 2$, then $f(2)$ is not in the table, so undefined. So, $f(g(x))$ is defined only for $x=0$ and $x=3$ with values $2$ and $-1$ respectively. 4. **Using the graph to evaluate (f \circ g)(4):** - From the graph, find $g(4)$: since $g(x)$ passes through $(5,2)$ and starts at $(0,0)$ with a less steep slope, approximate $g(4) \approx 1.6$. - Then find $f(g(4)) = f(1.6)$: $f(x)$ passes through $(0,0)$ and $(4,4)$, so $f(1.6) \approx 1.6$. So, $(f \circ g)(4) \approx 1.6$. 5. **Using the graph to evaluate (g \circ f)(3):** - From the graph, find $f(3)$: $f(3) \approx 3$ (since $f$ is linear passing through $(0,0)$ and $(4,4)$). - Then find $g(f(3)) = g(3)$: from the graph, $g(3) \approx 1.2$. So, $(g \circ f)(3) \approx 1.2$. **Final answers:** - (f \circ g)(6) = 5 - (f \circ f)(5) = 4 - $f(g(x))$ defined only for $x=0$ and $x=3$ with values 2 and -1 respectively - (f \circ g)(4) \approx 1.6 - (g \circ f)(3) \approx 1.2