1. **State the problem:** We are given a piecewise function: $$f(x) = \begin{cases} 3x + 5 & -4 \leq x \leq -1 \\ 2 & -1 \leq x < 3 \\ -x + 2 & 3 \leq x \leq 4 \end{cases}$$ We need to find the **range** of this function, i.e., all possible output values $f(x)$ can take. 2. **Analyze each piece separately:** - For $-4 \leq x \leq -1$, $f(x) = 3x + 5$. - Evaluate at endpoints: - At $x = -4$: $f(-4) = 3(-4) + 5 = -12 + 5 = -7$ - At $x = -1$: $f(-1) = 3(-1) + 5 = -3 + 5 = 2$ - Since $3x + 5$ is linear and increasing (slope 3 > 0), the range on this interval is all values from $-7$ to $2$ inclusive. - For $-1 \leq x < 3$, $f(x) = 2$ (constant). - The function value is always $2$ on this interval. - For $3 \leq x \leq 4$, $f(x) = -x + 2$. - Evaluate at endpoints: - At $x = 3$: $f(3) = -3 + 2 = -1$ - At $x = 4$: $f(4) = -4 + 2 = -2$ - Since slope is $-1$ (decreasing), the range on this interval is from $-1$ down to $-2$ inclusive. 3. **Combine all ranges:** - From first piece: $[-7, 2]$ - From second piece: $\{2\}$ (already included in first piece's range) - From third piece: $[-2, -1]$ 4. **Final range:** The union of these intervals is: $$[-7, 2] \cup [-2, -1] = [-7, 2]$$ Because $[-2, -1]$ is a subset of $[-7, 2]$, the overall range is $[-7, 2]$. **Answer:** The range of $f$ is $$\boxed{[-7, 2]}$$.