Piecewise functions worksheet with three problems to graph and evaluate.

Let's solve each of the piecewise functions step by step. We'll:

1. Graph each function manually (described in words since we can't draw here).
2. Evaluate the requested values.

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Problem 1

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

#### Step 1: Graphing
- For $ x < 0 $: Use $ f(x) = 2x + 3 $
- This is a line with slope 2 and y-intercept 3.
- But only for $ x < 0 $. So draw the line to the left of $ x = 0 $, open circle at $ x = 0 $.
- For $ x \geq 0 $: Use $ f(x) = 3 - x $
- This is a line with slope -1 and y-intercept 3.
- Draw this for $ x \geq 0 $, including $ x = 0 $, so closed circle at $ x = 0 $.

At $ x = 0 $:
- Left side: $ 2(0) + 3 = 3 $ → open circle at (0,3)
- Right side: $ 3 - 0 = 3 $ → closed circle at (0,3)

So both sides meet at (0,3), but the left side doesn't include it, right side does.

#### Evaluate:
- $ f(0) $: Since $ x = 0 \geq 0 $, use $ f(x) = 3 - x $
$ f(0) = 3 - 0 = \boxed{3} $

- $ f(-3) $: Since $ -3 < 0 $, use $ f(x) = 2x + 3 $
$ f(-3) = 2(-3) + 3 = -6 + 3 = \boxed{-3} $

- $ f(2) $: Since $ 2 \geq 0 $, use $ f(x) = 3 - x $
$ f(2) = 3 - 2 = \boxed{1} $

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Problem 2

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

#### Graphing:
- For $ x \leq 0 $: $ f(x) = x + 3 $
- Line with slope 1, y-intercept 3.
- Include $ x = 0 $: closed circle at (0,3)
- For $ 0 < x \leq 2 $: $ f(x) = 3 $
- Horizontal line at $ y = 3 $
- Open circle at $ x = 0 $ (since $ x > 0 $), closed circle at $ x = 2 $
- For $ x > 2 $: $ f(x) = 2x - 1 $
- Line with slope 2, y-intercept -1.
- Starts after $ x = 2 $, open circle at $ x = 2 $

Check continuity:
- At $ x = 0 $: left side gives $ f(0) = 0 + 3 = 3 $, middle piece starts at $ x > 0 $, so open at (0,3), but left includes (0,3). So there’s a closed dot on the left, open on the right — fine.
- At $ x = 2 $: middle gives $ f(2) = 3 $, right side starts at $ x > 2 $, so open at (2,3), but middle has closed dot at (2,3).

So graph has:
- Line from $ x \to -\infty $ to $ x = 0 $: $ y = x+3 $
- Horizontal line from $ x = 0^+ $ to $ x = 2 $: $ y = 3 $
- Then for $ x > 2 $: $ y = 2x - 1 $

#### Evaluate:
- $ f(-1) $: $ -1 \leq 0 $, so use $ f(x) = x + 3 $
$ f(-1) = -1 + 3 = \boxed{2} $

- $ f(1) $: $ 0 < 1 \leq 2 $, so use $ f(x) = 3 $
$ f(1) = \boxed{3} $

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Problem 3

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

Wait — notice: There is no definition for $ -2 < x \leq 2 $!
This is important.

So the function is defined only when:
- $ x \leq -2 $: $ f(x) = 3x + 5 $
- $ x > 2 $: $ f(x) = x - 4 $
- No rule for $ -2 < x \leq 2 $ → undefined in that interval.

So the graph consists of:
- A ray for $ x \leq -2 $: $ f(x) = 3x + 5 $
- Slope 3, y-intercept 5
- Closed circle at $ x = -2 $
- A ray for $ x > 2 $: $ f(x) = x - 4 $
- Slope 1, y-intercept -4
- Open circle at $ x = 2 $

But no graph between $ x = -2 $ and $ x = 2 $.

#### Evaluate:
- $ f(2) $: Is $ 2 \leq -2 $? No. Is $ 2 > 2 $? No. So undefined
$ f(2) = \boxed{\text{undefined}} $

- $ f(-3) $: $ -3 \leq -2 $, so use $ f(x) = 3x + 5 $
$ f(-3) = 3(-3) + 5 = -9 + 5 = \boxed{-4} $

- $ f(0) $: $ 0 $ is not $ \leq -2 $, nor $ > 2 $ → undefined
$ f(0) = \boxed{\text{undefined}} $

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Final Answers:

#### Problem 1:
- $ f(0) = \boxed{3} $
- $ f(-3) = \boxed{-3} $
- $ f(2) = \boxed{1} $

#### Problem 2:
- $ f(-1) = \boxed{2} $
- $ f(1) = \boxed{3} $

#### Problem 3:
- $ f(2) = \boxed{\text{undefined}} $
- $ f(-3) = \boxed{-4} $
- $ f(0) = \boxed{\text{undefined}} $

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Summary for Graphing (Instructions):

1. Problem 1: Two lines meeting at (0,3):
- Left: $ y = 2x + 3 $ for $ x < 0 $, open circle at (0,3)
- Right: $ y = 3 - x $ for $ x \geq 0 $, closed circle at (0,3)

2. Problem 2: Three segments:
- $ y = x + 3 $ for $ x \leq 0 $, closed at (0,3)
- $ y = 3 $ for $ 0 < x \leq 2 $, open at (0,3), closed at (2,3)
- $ y = 2x - 1 $ for $ x > 2 $, open at (2,3)

3. Problem 3: Two rays:
- $ y = 3x + 5 $ for $ x \leq -2 $, closed at (-2, -1)
- $ y = x - 4 $ for $ x > 2 $, open at (2, -2)
- Gap between $ x = -2 $ and $ x = 2 $

> ✔ Always check which piece applies based on the domain condition!

Let me know if you'd like a sketch description or help plotting points!