Piecewise Function Worksheet PDF | PDF - Free Printable

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 pieces meet at (0,3), but the left one is not defined at $ x = 0 $, so it’s an open circle on the left, closed on the right.

#### Step 2: Evaluate

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

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

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

✔ Answers:
- $ f(0) = 3 $
- $ f(-3) = -3 $
- $ f(2) = 1 $

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Problem 2

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

#### Step 1: 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 $
- Not defined at $ x = 0 $ → open circle at (0,3)
- Defined at $ x = 2 $ → closed circle at (2,3)
- For $ x > 2 $: $ f(x) = 2x - 1 $
- Line with slope 2, y-intercept -1
- Start after $ x = 2 $ → open circle at (2,3) because $ x > 2 $
- But wait: at $ x = 2 $, the middle piece gives $ f(2) = 3 $
- The third piece starts after $ x = 2 $, so open circle at (2,3) for this part
- But since $ f(2) = 3 $ from the middle piece, we have a closed circle at (2,3) from that segment

So:
- At $ x = 0 $: closed circle from first piece, open circle from second — but both give $ y = 3 $, so it's continuous there.
- At $ x = 2 $: closed circle from second piece, open circle from third — no issue.

#### Step 2: Evaluate

- $ f(-1) $: $ -1 \leq 0 $ → use $ f(x) = x + 3 $
$$
f(-1) = -1 + 3 = \boxed{2}
$$

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

✔ Answers:
- $ f(-1) = 2 $
- $ f(1) = 3 $

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Problem 3

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

Wait! There's a gap between $ x = -2 $ and $ x = 2 $. No definition for $ -2 < x \leq 2 $.

So:
- Defined only for $ x \leq -2 $ and $ x > 2 $
- Undefined for $ -2 < x \leq 2 $

#### Step 1: Graphing
- For $ x \leq -2 $: $ f(x) = 3x + 5 $
- Line with slope 3, y-intercept 5
- Include $ x = -2 $ → closed circle at $ x = -2 $
- For $ x > 2 $: $ f(x) = x - 4 $
- Line with slope 1, y-intercept -4
- Start after $ x = 2 $ → open circle at $ x = 2 $

But note: no function defined between $ x = -2 $ and $ x = 2 $, so gap in the graph.

#### Step 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 $ → use $ f(x) = 3x + 5 $
$$
f(-3) = 3(-3) + 5 = -9 + 5 = \boxed{-4}
$$

- $ f(0) $: $ 0 $ is not $ \leq -2 $ and not $ > 2 $ → undefined
$$
f(0) = \boxed{\text{undefined}}
$$

✔ Answers:
- $ f(2) = \text{undefined} $
- $ f(-3) = -4 $
- $ f(0) = \text{undefined} $

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✔ Final Answers Summary:

#### Problem 1:
- $ f(0) = 3 $
- $ f(-3) = -3 $
- $ f(2) = 1 $

#### Problem 2:
- $ f(-1) = 2 $
- $ f(1) = 3 $

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

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📝 Notes for Graphing (Visual Description):

- Problem 1: Two lines meeting at (0,3). Left side: increasing line ($2x+3$) up to open circle at (0,3). Right side: decreasing line ($3-x$) starting with closed circle at (0,3).

- Problem 2:
- Left: line $x+3$ from $-\infty$ to (0,3), closed circle.
- Middle: horizontal line $y=3$ from (0,3) open to (2,3) closed.
- Right: line $2x-1$ starting just after $x=2$, open circle at (2,3), going upward.

- Problem 3:
- Left: line $3x+5$ for $x \leq -2$, ending at closed point $(-2, f(-2)) = (-2, 3(-2)+5) = (-2, -1)$
- Then no graph from $x = -2$ to $x = 2$
- Then for $x > 2$: line $x - 4$, starting with open circle at $(2, 2-4) = (2, -2)$, then increasing.

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