Piecewise Function Worksheet PDF | PDF - Free Printable

Problem: Graph each piecewise function and evaluate the given points.

We will solve this step by step for each function.

---

Function 1:

$$
f(x) =
\begin{cases}
2x + 3, & x < 0 \\
3 - x, & x \geq 0
\end{cases}
$$

#### Step 1: Graph the function
- For \( x < 0 \): The function is \( f(x) = 2x + 3 \). This is a straight line with slope 2 and y-intercept 3.
- For \( x \geq 0 \): The function is \( f(x) = 3 - x \). This is a straight line with slope -1 and y-intercept 3.

The graph will have two segments:
1. A line \( y = 2x + 3 \) for \( x < 0 \).
2. A line \( y = 3 - x \) for \( x \geq 0 \).

At \( x = 0 \), the value of the function is \( f(0) = 3 - 0 = 3 \). So, the graph will have a point at \( (0, 3) \) for the second piece.

#### Step 2: Evaluate the function at the given points
1. Evaluate \( f(0) \):
- Since \( 0 \geq 0 \), use the second piece: \( f(x) = 3 - x \).
- \( f(0) = 3 - 0 = 3 \).

2. Evaluate \( f(-3) \):
- Since \( -3 < 0 \), use the first piece: \( f(x) = 2x + 3 \).
- \( f(-3) = 2(-3) + 3 = -6 + 3 = -3 \).

3. Evaluate \( f(2) \):
- Since \( 2 \geq 0 \), use the second piece: \( f(x) = 3 - x \).
- \( f(2) = 3 - 2 = 1 \).

#### Final Answers for Function 1:
$$
f(0) = 3, \quad f(-3) = -3, \quad f(2) = 1
$$

---

Function 2:

$$
f(x) =
\begin{cases}
x + 3, & x \leq 0 \\
3, & 0 < x \leq 2 \\
2x - 1, & x > 2
\end{cases}
$$

#### Step 1: Graph the function
- For \( x \leq 0 \): The function is \( f(x) = x + 3 \). This is a straight line with slope 1 and y-intercept 3.
- For \( 0 < x \leq 2 \): The function is a constant \( f(x) = 3 \). This is a horizontal line at \( y = 3 \).
- For \( x > 2 \): The function is \( f(x) = 2x - 1 \). This is a straight line with slope 2 and y-intercept -1.

The graph will have three segments:
1. A line \( y = x + 3 \) for \( x \leq 0 \).
2. A horizontal line \( y = 3 \) for \( 0 < x \leq 2 \).
3. A line \( y = 2x - 1 \) for \( x > 2 \).

#### Step 2: Evaluate the function at the given points
1. Evaluate \( f(-1) \):
- Since \( -1 \leq 0 \), use the first piece: \( f(x) = x + 3 \).
- \( f(-1) = -1 + 3 = 2 \).

2. Evaluate \( f(1) \):
- Since \( 0 < 1 \leq 2 \), use the second piece: \( f(x) = 3 \).
- \( f(1) = 3 \).

#### Final Answers for Function 2:
$$
f(-1) = 2, \quad f(1) = 3
$$

---

Function 3:

$$
f(x) =
\begin{cases}
3x + 5, & x \leq -2 \\
x - 4, & x > 2
\end{cases}
$$

#### Step 1: Graph the function
- For \( x \leq -2 \): The function is \( f(x) = 3x + 5 \). This is a straight line with slope 3 and y-intercept 5.
- For \( x > 2 \): The function is \( f(x) = x - 4 \). This is a straight line with slope 1 and y-intercept -4.

The graph will have two segments:
1. A line \( y = 3x + 5 \) for \( x \leq -2 \).
2. A line \( y = x - 4 \) for \( x > 2 \).

There is no defined behavior for \( -2 < x \leq 2 \), so the graph will have a gap in this interval.

#### Step 2: Evaluate the function at the given points
1. Evaluate \( f(2) \):
- Since \( 2 \) is not in the domain of either piece (it is outside both \( x \leq -2 \) and \( x > 2 \)), \( f(2) \) is undefined.

2. Evaluate \( f(-3) \):
- Since \( -3 \leq -2 \), use the first piece: \( f(x) = 3x + 5 \).
- \( f(-3) = 3(-3) + 5 = -9 + 5 = -4 \).

3. Evaluate \( f(0) \):
- Since \( 0 \) is not in the domain of either piece (it is outside both \( x \leq -2 \) and \( x > 2 \)), \( f(0) \) is undefined.

#### Final Answers for Function 3:
$$
f(2) = \text{undefined}, \quad f(-3) = -4, \quad f(0) = \text{undefined}
$$

---

Final Answers:

1. \( f(0) = 3, \quad f(-3) = -3, \quad f(2) = 1 \)
2. \( f(-1) = 2, \quad f(1) = 3 \)
3. \( f(2) = \text{undefined}, \quad f(-3) = -4, \quad f(0) = \text{undefined} \)

$$
\boxed{
\begin{aligned}
&f(0) = 3, \quad f(-3) = -3, \quad f(2) = 1 \\
&f(-1) = 2, \quad f(1) = 3 \\
&f(2) = \text{undefined}, \quad f(-3) = -4, \quad f(0) = \text{undefined}
\end{aligned}
}
$$