Question

which of the piecewise functions matches this graph?
$f(x)=\

$$\begin{cases}2x + 1 & \\text{if } x < 1 \\\\ -2x + 3 & \\text{if } x \\geq 1\\end{cases}$$

$
$f(x)=\

$$\begin{cases}x + 5 & \\text{if } x > -2 \\\\ -4 & \\text{if } x \\geq 2\\end{cases}$$

$
$f(x)=\

$$\begin{cases}-2x - 4 & \\text{if } x \\leq 2 \\\\ 4x - 9 & \\text{if } x > 2\\end{cases}$$

$
$f(x)=\

$$\begin{cases}x - 1 & \\text{if } x \\leq 2 \\\\ 2x - 1 & \\text{if } 2 < x \\leq 4 \\\\ 3x - 8 & \\text{if } x > 4\\end{cases}$$

$

Explanation:

To solve this, we analyze the graph and the piecewise functions:

Step 1: Analyze the Graph’s Components

The graph has two parts: a line (linear function) and a vertical segment (or a constant/linear part for \( x \geq \) some value). Let’s check the second function: \( f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

\). Wait, no—wait, the graph shows a line (slope, intercept) and a vertical-like segment? Wait, no, re-examining: the second function \( f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

\) has a line \( y = x + 5 \) (slope 1, y-intercept 5) for \( x < 2 \), and a horizontal line \( y = -4 \) for \( x \geq 2 \). But the graph’s line seems to have a negative slope? Wait, no—wait, the first function: \( f(x) =

$$\begin{cases} 2x + 1 & \text{if } x < 1 \\ -2x + 3 & \text{if } x \geq 1 \end{cases}$$

\). Let’s test \( x = 0 \): \( 2(0) + 1 = 1 \). \( x = 1 \): \( -2(1) + 3 = 1 \), so it’s continuous. But the graph’s line—wait, the second function: \( f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

\): at \( x = 0 \), \( y = 5 \); at \( x = 2 \), left limit is \( 2 + 5 = 7 \), right limit is \( -4 \), so discontinuous. But the graph’s line—wait, the correct one is likely the second function? Wait, no, let’s re-express. Wait, the graph shows a line (maybe \( y = -x + 5 \)?) and a vertical segment? No, the second function: \( f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

\). Wait, the graph’s line has a negative slope? Wait, no, the first function: \( 2x + 1 \) (slope 2) and \( -2x + 3 \) (slope -2). But the graph’s line—wait, the second function: \( x + 5 \) (slope 1) and \( -4 \) (horizontal). But the graph’s line—maybe the second function is correct? Wait, no, let’s check the options again.

Wait, the options are:

1. \( f(x) =

$$\begin{cases} 2x + 1 & \text{if } x < 1 \\ -2x + 3 & \text{if } x \geq 1 \end{cases}$$

\)

1. \( f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

\)

1. \( f(x) =

$$\begin{cases} -2x - 4 & \text{if } x \leq 2 \\ 4x - 9 & \text{if } x > 2 \end{cases}$$

\) (wait, no, the third function is \( f(x) =

$$\begin{cases} -2x - 4 & \text{if } x \leq 2 \\ 4x - 9 & \text{if } x > 2 \end{cases}$$

\))

1. \( f(x) =

$$\begin{cases} x - 1 & \text{if } x \leq 2 \\ 2x - 1 & \text{if } 2 < x \leq 4 \\ 3x - 8 & \text{if } x > 4 \end{cases}$$

\)

Wait, the graph shows a line (maybe \( y = x + 5 \)) and a horizontal line \( y = -4 \) for \( x \geq 2 \). So the second function: \( f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

\) matches the graph’s line (slope 1, y-intercept 5) for \( x < 2 \) and a horizontal segment at \( y = -4 \) for \( x \geq 2 \).

Answer:

The piecewise function that matches the graph is \( \boldsymbol{f(x) =

$$\begin{cases} x + 5 & \text{if } x < 2 \\ -4 & \text{if } x \geq 2 \end{cases}$$

} \) (the second option in the list).