Question

lesson objective
students will evaluate piecewise functions by determining which rule
the piecewise function (problems 1–4)
$f(x)=\

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

$
problems (evaluate)

1. $f(-4) = $
2. $f(-2) = $
3. $f(0) = $
4. $f(2) = $

Explanation:

Problem 1: Evaluate \( f(-4) \)

Step 1: Determine the applicable rule

Since \( -4 \leq -2 \), we use the rule \( f(x) = x + 3 \).

Step 2: Substitute \( x = -4 \) into the rule

\( f(-4) = -4 + 3 \)

Step 3: Calculate the result

\( -4 + 3 = -1 \)

Step 1: Determine the applicable rule

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

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

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

Step 3: Calculate the result

\( -2 + 3 = 1 \)

Step 1: Determine the applicable rule

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

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

\( f(0) = 0^2 \)

Step 3: Calculate the result

\( 0^2 = 0 \)

Answer:

\( -1 \)

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