Question

type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
the graph represents the piecewise function:
$f(x)=\

$$\begin{cases}\\square, \\text{ if } -3 \\leq x < -1\\\\\\square, \\text{ if } -1 \\leq x \\leq 1\\end{cases}$$

$

Explanation:

Step1: Analyze the first piece ($-3 \leq x < -1$)

The graph for $-3 \leq x < -1$ is a horizontal line? Wait, no, looking at the graph, from $x = -3$ (closed dot) to $x = -1$ (open dot), wait no, wait the first segment: when $-3 \leq x < -1$? Wait, no, the graph: at $x = -3$, there's a closed dot, and then a line to $x = -1$ (open dot) at $y = 2$? Wait no, wait the graph: let's check the coordinates. Wait, the leftmost part: from $x = -6$ (or whatever) to $x = -3$, it's a horizontal line at $y = 0$? Wait no, the graph has a segment from $(-3, 0)$ (closed dot) to $(-1, 2)$ (open dot). Wait, slope: $\frac{2 - 0}{-1 - (-3)} = \frac{2}{2} = 1$. So the equation is $y - 0 = 1(x + 3)$, so $y = x + 3$. Wait, but wait, when $-3 \leq x < -1$, the graph is a line from $(-3, 0)$ to $(-1, 2)$. Then, for $-1 \leq x \leq 1$, the graph: at $x = -1$, closed dot at $y = 5$? Wait no, looking at the graph: from $x = -1$ (closed dot) to $x = 1$ (closed dot) at $y = 5$? Wait, the horizontal line at $y = 5$ from $x = -1$ to $x = 1$? Wait, no, the graph: the top segment is from $x = -2$ (closed dot) to $x = 1$ (closed dot) at $y = 5$? Wait, maybe I misread. Wait, the problem is a piecewise function with two pieces: first for $-3 \leq x < -1$, second for $-1 \leq x \leq 1$.

Wait, let's re-examine the graph:

1. For the interval $-3 \leq x < -1$: The graph is a line segment starting at $(-3, 0)$ (closed dot) and ending at $(-1, 2)$ (open dot). So the slope is $\frac{2 - 0}{-1 - (-3)} = 1$. So the equation is $y = x + 3$ (since when $x = -3$, $y = 0$: $-3 + 3 = 0$; when $x = -1$, $y = -1 + 3 = 2$, which matches the open dot at $(-1, 2)$).

1. For the interval $-1 \leq x \leq 1$: The graph is a horizontal line at $y = 5$, from $x = -1$ (closed dot) to $x = 1$ (closed dot). So the equation is $y = 5$.

Wait, but let's confirm:

• For $-3 \leq x < -1$: The line goes from $(-3, 0)$ to $(-1, 2)$. So slope $m = (2 - 0)/(-1 - (-3)) = 2/2 = 1$. So equation: $y = x + 3$.

• For $-1 \leq x \leq 1$: The horizontal line at $y = 5$, so $y = 5$.

Wait, but let's check the first piece: when $x = -3$, $y = -3 + 3 = 0$, correct (closed dot). When $x = -1$, $y = -1 + 3 = 2$, which is the open dot, correct. Then, for the second piece: $-1 \leq x \leq 1$, the graph is horizontal at $y = 5$, so $y = 5$.

So the piecewise function is:

$f(x) =

$$\begin{cases} x + 3 & \text{if } -3 \leq x < -1 \\ 5 & \text{if } -1 \leq x \leq 1 \end{cases}$$

$

Wait, but let's confirm with the graph. The first segment: from $(-3, 0)$ to $(-1, 2)$: yes, $y = x + 3$. The second segment: from $(-1, 5)$ to $(1, 5)$: horizontal line $y = 5$.

So the first box (for $-3 \leq x < -1$) is $x + 3$, and the second box (for $-1 \leq x \leq 1$) is $5$.

Wait, but let's check the slope again. From $(-3, 0)$ to $(-1, 2)$: change in y is 2, change in x is 2, so slope 1. So equation is $y = 1(x + 3) + 0 = x + 3$. Correct.

For the second interval, $-1 \leq x \leq 1$, the graph is a horizontal line at $y = 5$, so $f(x) = 5$.

Step1: Find the first piece ($-3 \leq x < -1$)

The segment goes from $(-3, 0)$ (closed) to $(-1, 2)$ (open). Slope $m = \frac{2 - 0}{-1 - (-3)} = 1$. Using point-slope form: $y - 0 = 1(x + 3) \implies y = x + 3$.

Step2: Find the second piece ($-1 \leq x \leq 1$)

The segment is horizontal at $y = 5$ (from $x = -1$ to $x = 1$, both closed dots), so $f(x) = 5$.

Answer:

First box: $x + 3$
Second box: $5$