Question

1. $f(x)=\

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

$

Explanation:

Step1: Analyze the first piece of the function

For \( f(x)=-x + 3\) when \(x<2\).

• Find two points:
• When \(x = 0\), \(f(0)=-0 + 3=3\). So the point is \((0,3)\).
• When \(x=1\), \(f(1)=-1 + 3 = 2\). So the point is \((1,2)\).
• Since \(x<2\), the line \(y=-x + 3\) for this part has an open - circle at \(x = 2\). When \(x = 2\), \(y=-2+3 = 1\).

Step2: Analyze the second piece of the function

For \(f(x)=2x-3\) when \(x\geq2\).

• Find two points:
• When \(x = 2\), \(f(2)=2\times2-3=1\). So the point is \((2,1)\) (closed - circle because \(x = 2\) is included).
• When \(x=3\), \(f(3)=2\times3-3=3\). So the point is \((3,3)\).

Step3: Plot the points and draw the lines

• Plot \((0,3)\), \((1,2)\) for \(y=-x + 3,x<2\) and connect them with a line (with an open - circle at \(x = 2\)).
• Plot \((2,1)\) and \((3,3)\) for \(y = 2x-3,x\geq2\) and connect them with a line (with a closed - circle at \(x = 2\)).

Answer:

The graph of the piece - wise function \(f(x)=

$$\begin{cases}-x + 3&x<2\\2x-3&x\geq2\end{cases}$$

\) is composed of two line segments: one for \(y=-x + 3\) (with an open - circle at \(x = 2\)) and one for \(y=2x-3\) (with a closed - circle at \(x = 2\)).