Question

use the image to answer the question.
create a piecewise function that represents the graph.
(1 point)
\\( \circ f(x)=\

$$\begin{cases}1, & -3\\leq x < -1 \\\\ 2x - 1, & -1\\leq x \\leq 1\\end{cases}$$

\\)
\\( \circ f(x)=\

$$\begin{cases}1, & -3\\leq x < -1 \\\\ 2x, & -1\\leq x \\leq 1\\end{cases}$$

\\)
\\( \circ f(x)=\

$$\begin{cases}x + 1, & -3 < x < -1 \\\\ 2x - 1, & -1\\leq x \\leq 1\\end{cases}$$

\\)
\\( \circ f(x)=\

$$\begin{cases}1, & -3 < x \\leq -1 \\\\ 2x - 1, & -1 < x < 1\\end{cases}$$

\\)

Explanation:

Step1: Analyze the first piece (horizontal line)

The horizontal line has a solid dot at \( x = -3 \) and an open dot at \( x = -1 \), with \( y = 1 \). So the domain for this piece is \( -3 \leq x < -1 \) and the function is \( f(x)=1 \).

Step2: Analyze the second piece (linear line)

The linear part: let's check the points. At \( x = -1 \), the open dot from the first piece, and at \( x = 1 \), a solid dot. Let's test the function \( 2x - 1 \). At \( x = 1 \), \( 2(1)-1 = 1 \), which matches the solid dot. At \( x = 0 \), \( 2(0)-1=-1 \), which matches the dashed line. The domain here is \( -1 \leq x \leq 1 \) (since solid dots at both ends) and the function is \( 2x - 1 \).

Step3: Match with options

Now check the options:

• First option: \( f(x)=

$$\begin{cases}1, & -3\leq x < -1 \\ 2x - 1, & -1\leq x \leq 1\end{cases}$$

\) matches our analysis.

• Second option: \( 2x \) at \( x = 1 \) gives \( 2 \), but the graph has \( y = 1 \) at \( x = 1 \), so wrong.
• Third option: First piece is \( x + 1 \), at \( x = -3 \), \( -3 + 1=-2

eq1 \), wrong.

• Fourth option: Domain for second piece is \( -1 < x < 1 \), but the graph has a solid dot at \( x = 1 \), so wrong.

Answer:

\( f(x)=

$$\begin{cases}1, & -3\leq x < -1 \\ 2x - 1, & -1\leq x \leq 1\end{cases}$$

\) (the first option among the given choices)