Question

lesson objective
students will evaluate piecewise functions by identifying which rule applie substituting to compute the output.
the piecewise function (problems 1–3)
$f(x)=\

$$\begin{cases}x - 1, & x\\leq0\\\\x + 2, & 0 < x < 3\\\\2x - 5, & x\\geq3\\end{cases}$$

$
problems (evaluate)

1. $f(-2) = $
2. $f(0) = $
3. $f(2) = $

the piecewise function (problems 4–6)
$f(x)=\

$$\begin{cases}x^2, & x < -1\\\\x + 1, & -1\\leq x\\leq2\\\\4, & x > 2\\end{cases}$$

$
problems (evaluate)

1. $f(-3) = $
2. $f(-1) = $
3. $f(3) = $

Explanation:

Problem 1: Evaluate \( f(-2) \)

Step 1: Identify the rule for \( x = -2 \)

Since \( -2 \leq 0 \), we use the rule \( f(x)=x - 1 \).

Step 2: Substitute \( x=-2 \) into the rule

\( f(-2)=-2 - 1=-3 \)

Step 1: Identify the rule for \( x = 0 \)

Since \( 0\leq0 \), we use the rule \( f(x)=x - 1 \).

Step 2: Substitute \( x = 0 \) into the rule

\( f(0)=0 - 1=-1 \)

Step 1: Identify the rule for \( x = 2 \)

Since \( 0<2<3 \), we use the rule \( f(x)=x + 2 \).

Step 2: Substitute \( x = 2 \) into the rule

\( f(2)=2 + 2=4 \)

Answer:

\( -3 \)

Problem 2: Evaluate \( f(0) \)