1. **State the problem:** We are given the absolute value function $$f(x) = |x + 3|$$ and need to express it as a piecewise function. 2. **Recall the definition of absolute value:** For any expression $$A$$, $$|A| = \begin{cases} A & \text{if } A \geq 0 \\ -A & \text{if } A < 0 \end{cases}$$ 3. **Apply this to $$f(x) = |x + 3|$$:** Here, $$A = x + 3$$. 4. **Determine the critical point:** The expression inside the absolute value changes sign at $$x + 3 = 0 \Rightarrow x = -3$$. 5. **Write the piecewise function:** $$ f(x) = \begin{cases} -(x + 3) & \text{if } x + 3 < 0 \\ x + 3 & \text{if } x + 3 \geq 0 \end{cases} $$ which simplifies to $$ f(x) = \begin{cases} -x - 3 & \text{if } x < -3 \\ x + 3 & \text{if } x \geq -3 \end{cases} $$ 6. **Interpret the options:** The correct piece for $$x \geq -3$$ is $$x + 3$$. **Final answer:** The piecewise function includes the piece $$x + 3$$ for $$x \geq -3$$. This matches the option: ○ x + 3, x ≥ -3