1. **Problem statement:** Given the function $k(x) = |x - 3| - |x + 5|$, we need to write it as a simplified piecewise function by considering three intervals. 2. **Intervals to consider:** The absolute value expressions change behavior at points where their inside is zero: $x=3$ and $x=-5$. So, the intervals are: - $x < -5$ - $-5 \leq x < 3$ - $x \geq 3$ 3. **Recall the definition of absolute value:** - For $|a|$, if $a \geq 0$, then $|a| = a$. - If $a < 0$, then $|a| = -a$. 4. **Evaluate $k(x)$ on each interval:** - For $x < -5$: - $x - 3 < 0$ so $|x - 3| = -(x - 3) = -x + 3$ - $x + 5 < 0$ so $|x + 5| = -(x + 5) = -x - 5$ - Therefore, $k(x) = (-x + 3) - (-x - 5) = -x + 3 + x + 5 = 8$ - For $-5 \leq x < 3$: - $x - 3 < 0$ so $|x - 3| = -(x - 3) = -x + 3$ - $x + 5 \geq 0$ so $|x + 5| = x + 5$ - Therefore, $k(x) = (-x + 3) - (x + 5) = -x + 3 - x - 5 = -2x - 2$ - For $x \geq 3$: - $x - 3 \geq 0$ so $|x - 3| = x - 3$ - $x + 5 \geq 0$ so $|x + 5| = x + 5$ - Therefore, $k(x) = (x - 3) - (x + 5) = x - 3 - x - 5 = -8$ 5. **Piecewise function:** $$ k(x) = \begin{cases} 8 & x < -5 \\ -2x - 2 & -5 \leq x < 3 \\ -8 & x \geq 3 \end{cases} $$ 6. **Domain:** The function is defined for all real numbers since absolute values are defined everywhere. 7. **Range:** - On $x < -5$, $k(x) = 8$ (constant) - On $-5 \leq x < 3$, $k(x) = -2x - 2$ is linear decreasing from $k(-5) = -2(-5) - 2 = 10 - 2 = 8$ to $k(3) = -2(3) - 2 = -6 - 2 = -8$ - On $x \geq 3$, $k(x) = -8$ (constant) So the range is all values between $-8$ and $8$, inclusive. 8. **Summary:** - Domain: $(-\infty, \infty)$ - Range: $[-8, 8]$