Question

12.
$f(x)=\

$$\begin{cases}\\frac{2}{3}x + 3, & x \\leq 0 \\\\ 3, & 0 < x < 2 \\\\ -\\frac{1}{2}x, & x \\geq 2 \\end{cases}$$

$

Explanation:

Step1: Analyze the first piece \(y = \frac{2}{3}x+3\) (\(x\leq0\))

• When \(x = 0\), \(y=\frac{2}{3}\times0 + 3=3\).
• Another point: when \(x=-3\), \(y=\frac{2}{3}\times(-3)+3=-2 + 3=1\). Plot the line segment for \(x\leq0\) using the two - point \((0,3)\) and \((-3,1)\).

Step2: Analyze the second piece \(y = 3\) (\(0\lt x\lt2\))

• This is a horizontal line segment. The endpoints are not included. We can consider open - circles at \(x = 0\) and \(x = 2\) with \(y = 3\).

Step3: Analyze the third piece \(y=-\frac{1}{2}x\) (\(x\geq2\))

• When \(x = 2\), \(y=-\frac{1}{2}\times2=-1\).
• When \(x = 4\), \(y=-\frac{1}{2}\times4=-2\). Plot the line segment for \(x\geq2\) using the point \((2,-1)\) and \((4,-2)\).

Answer:

Plot the three - piece function:

1. For \(y=\frac{2}{3}x + 3\) (\(x\leq0\)), use the points \((-3,1)\) and \((0,3)\) (closed - circle at \((0,3)\)).
2. For \(y = 3\) (\(0\lt x\lt2\)), it's a horizontal line with open - circles at \(x = 0\) and \(x = 2\).
3. For \(y=-\frac{1}{2}x\) (\(x\geq2\)), use the points \((2,-1)\) and \((4,-2)\) (closed - circle at \((2,-1)\)).