1. **State the problem:** We need to analyze the function $$f(x) = p|x + 2| - |2x - 4|$$ where $p$ is a constant. 2. **Recall the absolute value definition:** For any real number $a$, $$|a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases}$$ 3. **Identify critical points:** The expressions inside the absolute values change sign at points where their arguments are zero. - For $|x + 2|$, zero at $x = -2$. - For $|2x - 4|$, zero at $2x - 4 = 0 \Rightarrow x = 2$. 4. **Split the domain into intervals based on these points:** - Interval 1: $x < -2$ - Interval 2: $-2 \leq x < 2$ - Interval 3: $x \geq 2$ 5. **Evaluate $f(x)$ on each interval:** - For $x < -2$: $$|x + 2| = -(x + 2) = -x - 2$$ $$|2x - 4| = -(2x - 4) = -2x + 4$$ So, $$f(x) = p(-x - 2) - (-2x + 4) = -px - 2p + 2x - 4 = (-p + 2)x - 2p - 4$$ - For $-2 \leq x < 2$: $$|x + 2| = x + 2$$ $$|2x - 4| = -(2x - 4) = -2x + 4$$ So, $$f(x) = p(x + 2) - (-2x + 4) = px + 2p + 2x - 4 = (p + 2)x + 2p - 4$$ - For $x \geq 2$: $$|x + 2| = x + 2$$ $$|2x - 4| = 2x - 4$$ So, $$f(x) = p(x + 2) - (2x - 4) = px + 2p - 2x + 4 = (p - 2)x + 2p + 4$$ 6. **Summary:** $$ f(x) = \begin{cases} (-p + 2)x - 2p - 4 & x < -2 \\ (p + 2)x + 2p - 4 & -2 \leq x < 2 \\ (p - 2)x + 2p + 4 & x \geq 2 \end{cases} $$ This piecewise linear function depends on $p$ and the intervals defined by the absolute value expressions. **Final answer:** The function $f(x)$ is piecewise linear as above, with breakpoints at $x = -2$ and $x = 2$.