Question

defining piecewise functions which defines the piecewise function shown? $f(x)=\

$$\begin{cases}-x - 2, & x < 0\\\\ -\\frac{x}{2}, & x \\geq 0\\end{cases}$$

$ $f(x)=\

$$\begin{cases}-x - 2, & x < 0\\\\ \\frac{x}{2}, & x \\geq 0\\end{cases}$$

$ $f(x)=\

$$\begin{cases}\\frac{x}{2}, & x < 0\\\\ x - 2, & x \\geq 0\\end{cases}$$

$

Explanation:

Step1: Analyze upper line (x < 0)

First, identify the line for \(x < 0\): it has a y-intercept at \(y=-2\) and slope \(m=1\). Using slope-intercept form \(y=mx+b\), we get \(y = 1x - 2 = x - 2\). Wait, no—wait, check points: when \(x=-2\), the open circle is at \(y=0\): \(0 = -(-2) - 2 = 2-2=0\), correct. When \(x=0\), the open circle would be \(y=-0-2=-2\), which matches. So the upper line (x<0) is \(f(x) = x - 2\)? No, wait slope: from (-2,0) to (0,-2): slope is \(\frac{-2-0}{0-(-2)}=\frac{-2}{2}=-1\). So \(y = -x -2\). For \(x < 0\), open circle at \(x=-2\), \(y=0\): \(0=-(-2)-2=0\), correct.

Step2: Analyze lower line (x ≥ 0)

Now the line for \(x \geq 0\): closed point at (0,0), slope is \(\frac{\Delta y}{\Delta x}=\frac{2}{1}=2\)? No, wait: when \(x=2\), \(y=1\), so slope \(m=\frac{1}{2}\). Equation: \(y=\frac{1}{2}x\). Check \(x=0\): \(y=0\), matches the closed point. When \(x=4\), \(y=2\), which fits \(\frac{4}{2}=2\).

Step3: Match to options

Now match the pieces: for \(x < 0\), \(f(x)=-x-2\); for \(x \geq 0\), \(f(x)=\frac{x}{2}\). This matches the middle option.

Answer:

$f(x)=

$$\begin{cases} -x-2, & x < 0 \\ \frac{x}{2}, & x \geq 0 \end{cases}$$

$