Question: For each of the functions listed below in parts a to d, determine if the function has a range of $(-\infty, 4)$.\na. $f(x) = \begin{cases} x - 3, & \text{if } x < -2 \\ 5 - x, & \text{if } x > 1 \end{cases}$\nNo\nYes\nb. $h(x) = \begin{cases} x - 1, & \text{if } x < -1 \\ 3 - x, & \text{if } x > 2 \end{cases}$\nNo\nYes\nGraph shape/math: a and b are piecewise linear functions with two disconnected line segments each; part a uses a left branch with slope 1 for $x < -2$ and a right branch with slope $-1$ for $x > 1$, and part b uses a left branch with slope 1 for $x < -1$ and a right branch with slope $-1$ for $x > 2$. position_hint = top-left
1. **State the problem:** Determine if the range of each piecewise function $f(x)$ and $h(x)$ is $(-\infty, 4)$.\n\n2. **Recall:** The range of a function is the set of all possible output values. We analyze each piecewise branch separately and then combine their ranges.\n\n### Part a: $f(x) = \begin{cases} x - 3, & x < -2 \\ 5 - x, & x > 1 \end{cases}$\n3. For the left branch $x - 3$ with $x < -2$:\n- As $x \to -\infty$, $x - 3 \to -\infty$.\n- At $x = -2$, $f(-2) = -2 - 3 = -5$ (not included since $x < -2$).\n- So the left branch outputs values less than $-5$. Range left branch: $(-\infty, -5)$.\n\n4. For the right branch $5 - x$ with $x > 1$:\n- As $x \to 1^+$, $5 - x \to 4$ (values just less than 4).\n- As $x \to \infty$, $5 - x \to -\infty$.\n- So the right branch outputs values from just below 4 down to $-\infty$. Range right branch: $(-\infty, 4)$.\n\n5. Combine ranges:\n- Left branch: $(-\infty, -5)$\n- Right branch: $(-\infty, 4)$\n- Union: $(-\infty, 4)$\n\n6. **Conclusion for part a:** The range of $f(x)$ is $(-\infty, 4)$, so answer is **Yes**.\n\n### Part b: $h(x) = \begin{cases} x - 1, & x < -1 \\ 3 - x, & x > 2 \end{cases}$\n7. For the left branch $x - 1$ with $x < -1$:\n- As $x \to -\infty$, $x - 1 \to -\infty$.\n- At $x = -1$, $h(-1) = -1 - 1 = -2$ (not included).\n- So left branch outputs values less than $-2$. Range left branch: $(-\infty, -2)$.\n\n8. For the right branch $3 - x$ with $x > 2$:\n- As $x \to 2^+$, $3 - x \to 1$ (values just less than 1).\n- As $x \to \infty$, $3 - x \to -\infty$.\n- So right branch outputs values from just below 1 down to $-\infty$. Range right branch: $(-\infty, 1)$.\n\n9. Combine ranges:\n- Left branch: $(-\infty, -2)$\n- Right branch: $(-\infty, 1)$\n- Union: $(-\infty, 1)$\n\n10. Since the combined range is $(-\infty, 1)$, which is a subset of $(-\infty, 4)$ but does not reach values close to 4, the range is not exactly $(-\infty, 4)$.\n\n11. **Conclusion for part b:** The range of $h(x)$ is $(-\infty, 1)$, so answer is **No**.