Function Table 005C42

1. **State the problem:** We need to complete the table for $f(x) = \frac{1}{3} x + 2$ and $g(x) = 2^{x + 3}$ at $x = -3, -2, 0$ and then determine where $f(x) = g(x)$ has a solution based on the table. 2. **Calculate $f(x)$ values:** - For $x = -3$: $$f(-3) = \frac{1}{3}(-3) + 2 = -1 + 2 = 1$$ - For $x = -2$: $$f(-2) = \frac{1}{3}(-2) + 2 = -\frac{2}{3} + 2 = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}$$ - For $x = 0$: $$f(0) = \frac{1}{3}(0) + 2 = 0 + 2 = 2$$ 3. **Calculate $g(x)$ values:** - For $x = -3$: $$g(-3) = 2^{-3 + 3} = 2^0 = 1$$ - For $x = -2$: $$g(-2) = 2^{-2 + 3} = 2^1 = 2$$ - For $x = 0$: $$g(0) = 2^{0 + 3} = 2^3 = 8$$ 4. **Complete the table:** \begin{array}{c|c|c} x & f(x) & g(x) \\\hline -4 & \frac{2}{3} & \frac{1}{2} \\ -3 & 1 & 1 \\ -2 & \frac{4}{3} & 2 \\ -1 & \frac{5}{3} & 4 \\ 0 & 2 & 8 \\ \end{array} 5. **Determine where $f(x) = g(x)$ has a solution:** - At $x = -4$, $f(x) = \frac{2}{3} \approx 0.67$ and $g(x) = \frac{1}{2} = 0.5$, so $f(x) > g(x)$. - At $x = -3$, $f(x) = 1$ and $g(x) = 1$, so $f(x) = g(x)$ exactly at $x = -3$. Therefore, the equation $f(x) = g(x)$ has a solution at $x = -3$. **Final answer:** - Completed table values as above. - Solution to $f(x) = g(x)$ is at $x = -3$.