1. **State the problem:** We need to sketch the graph of the function $$f(x) = |x + 2| - |x|$$ and find its range. 2. **Understand the absolute value function:** Recall that $$|a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases}$$ 3. **Analyze the function in pieces:** The function involves two absolute values, so we consider the critical points where the expressions inside the absolute values change sign: at $$x = -2$$ and $$x = 0$$. 4. **For $$x \geq 0$$:** - $$|x + 2| = x + 2$$ (since $$x + 2 \geq 2 > 0$$) - $$|x| = x$$ - So, $$f(x) = (x + 2) - x = 2$$ 5. **For $$-2 \leq x < 0$$:** - $$|x + 2| = x + 2$$ (since $$x + 2 \geq 0$$) - $$|x| = -x$$ (since $$x < 0$$) - So, $$f(x) = (x + 2) - (-x) = x + 2 + x = 2x + 2$$ 6. **For $$x < -2$$:** - $$|x + 2| = -(x + 2)$$ (since $$x + 2 < 0$$) - $$|x| = -x$$ (since $$x < 0$$) - So, $$f(x) = -(x + 2) - (-x) = -x - 2 + x = -2$$ 7. **Summarize the piecewise function:** $$ f(x) = \begin{cases} -2 & x < -2 \\ 2x + 2 & -2 \leq x < 0 \\ 2 & x \geq 0 \end{cases} $$ 8. **Graph description:** - For $$x < -2$$, $$f(x) = -2$$ (horizontal line) - For $$-2 \leq x < 0$$, $$f(x)$$ is a line with slope 2 and intercept 2 - For $$x \geq 0$$, $$f(x) = 2$$ (horizontal line) 9. **Find the range:** - At $$x = -2$$, $$f(-2) = 2(-2) + 2 = -4 + 2 = -2$$ - At $$x = 0$$, $$f(0) = 2(0) + 2 = 2$$ - On $$[-2,0]$$, $$f(x)$$ increases linearly from $$-2$$ to $$2$$ - For $$x < -2$$, $$f(x) = -2$$ - For $$x \geq 0$$, $$f(x) = 2$$ Thus, the range is $$[-2, 2]$$. **Final answer:** The function $$f(x)$$ is piecewise linear with values $$-2$$ for $$x < -2$$, increases linearly from $$-2$$ to $$2$$ on $$[-2,0]$$, and is constant $$2$$ for $$x \geq 0$$. The range of $$f$$ is $$[-2, 2]$$.