Piecewise Function Worksheet - Fill and Sign Printable Template Online - Free Printable

Let's solve each of the three piecewise functions step by step. We'll analyze each function, determine its domain and continuity, evaluate specific values, and sketch the graph.

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Problem 1:

$$
f(x) =
\begin{cases}
x - 8 & \text{if } x < -2 \\
x^2 - 2x - 3 & \text{if } x \geq -2
\end{cases}
$$

#### Step 1: Domain
- The function is defined for all real numbers:
- $ x < -2 $: linear expression
- $ x \geq -2 $: quadratic expression
- So, Domain: All real numbers → $ (-\infty, \infty) $

✔ Function: Yes
✔ Continuous: ? (We’ll check later)

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#### Step 2: Evaluate Function Values

(a) $ f(-4) $:
Since $ -4 < -2 $, use $ f(x) = x - 8 $:
$$
f(-4) = -4 - 8 = -12
$$

(b) $ f(0) $:
Since $ 0 \geq -2 $, use $ f(x) = x^2 - 2x - 3 $:
$$
f(0) = 0^2 - 2(0) - 3 = -3
$$

(c) $ f(2) $:
$ 2 \geq -2 $, so again use the quadratic:
$$
f(2) = 2^2 - 2(2) - 3 = 4 - 4 - 3 = -3
$$

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#### Step 3: Check Continuity at $ x = -2 $

To check continuity at $ x = -2 $, we need:

1. $ f(-2) $ exists → yes, since $ x = -2 \geq -2 $, so use quadratic:
$$
f(-2) = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5
$$

2. Left-hand limit as $ x \to -2^- $:
Use $ f(x) = x - 8 $:
$$
\lim_{x \to -2^-} f(x) = -2 - 8 = -10
$$

3. Right-hand limit as $ x \to -2^+ $:
Use $ f(x) = x^2 - 2x - 3 $:
$$
\lim_{x \to -2^+} f(x) = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5
$$

Now, compare:
- Left limit = $-10$, Right limit = $5$, $ f(-2) = 5 $
- Since left and right limits are not equal, the function is not continuous at $ x = -2 $

✘ Continuous: No

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#### Graph Sketching Tips:
- For $ x < -2 $: line $ y = x - 8 $, open circle at $ x = -2 $, value would be $-10$
- For $ x \geq -2 $: parabola $ y = x^2 - 2x - 3 $, closed circle at $ x = -2 $, value = 5
- There’s a jump discontinuity at $ x = -2 $

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✔ Summary for Problem 1:

- Domain: Yes (all real numbers)
- Function: Yes
- Continuous: No
- $ f(-4) = -12 $
- $ f(0) = -3 $
- $ f(2) = -3 $

---

Problem 2:

$$
f(x) =
\begin{cases}
2x - 1 & \text{if } x < 3 \\
x^2 - 3 & \text{if } x > 3
\end{cases}
$$

Note: No definition at $ x = 3 $ — this is important!

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#### Step 1: Domain
- Defined for $ x < 3 $ and $ x > 3 $
- Not defined at $ x = 3 $
- So, Domain: $ (-\infty, 3) \cup (3, \infty) $

✔ Function: Yes (each input has one output, even though not defined at 3)
✘ Continuous: No (not even defined at $ x = 3 $, so can't be continuous there)

---

#### Step 2: Evaluate Function Values

(a) $ f(-2) $:
$ -2 < 3 $, so use $ 2x - 1 $:
$$
f(-2) = 2(-2) - 1 = -4 - 1 = -5
$$

(b) $ f(0) $:
$ 0 < 3 $, so same rule:
$$
f(0) = 2(0) - 1 = -1
$$

(c) $ f(3) $:
Not defined! Because $ x = 3 $ is excluded from both cases.
So, $ f(3) = \boxed{\text{undefined}} $

---

#### Step 3: Graph Sketching Tips
- For $ x < 3 $: line $ y = 2x - 1 $, open circle at $ x = 3 $
- For $ x > 3 $: parabola $ y = x^2 - 3 $, open circle at $ x = 3 $
- At $ x = 3 $: hole in the graph

Check limits:
- Left limit: $ \lim_{x \to 3^-} f(x) = 2(3) - 1 = 5 $
- Right limit: $ \lim_{x \to 3^+} f(x) = 3^2 - 3 = 9 - 3 = 6 $
- Since left ≠ right, and function undefined at 3 → discontinuous

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✔ Summary for Problem 2:

- Domain: No (missing $ x = 3 $)
- Function: Yes
- Continuous: No
- $ f(-2) = -5 $
- $ f(0) = -1 $
- $ f(3) = \text{undefined} $

---

Problem 3:

$$
f(x) =
\begin{cases}
2x - 1 & \text{if } x \leq 2 \\
2x - 6 & \text{if } x > 2
\end{cases}
$$

---

#### Step 1: Domain
- Defined for $ x \leq 2 $ and $ x > 2 $
- So covers all real numbers → $ (-\infty, \infty) $

✔ Function: Yes
✔ Continuous: ? Let's check at $ x = 2 $

---

#### Step 2: Evaluate Function Values

(a) $ f(-4) $:
$ -4 \leq 2 $, so use $ 2x - 1 $:
$$
f(-4) = 2(-4) - 1 = -8 - 1 = -9
$$

(b) $ f(0) $:
$ 0 \leq 2 $, so same:
$$
f(0) = 2(0) - 1 = -1
$$

(c) $ f(2) $:
$ x = 2 \leq 2 $, so use $ 2x - 1 $:
$$
f(2) = 2(2) - 1 = 4 - 1 = 3
$$

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#### Step 3: Check Continuity at $ x = 2 $

- $ f(2) = 3 $
- Left-hand limit ($ x \to 2^- $): use $ 2x - 1 $ → $ 2(2) - 1 = 3 $
- Right-hand limit ($ x \to 2^+ $): use $ 2x - 6 $ → $ 2(2) - 6 = 4 - 6 = -2 $

So:
- Left limit = 3
- Right limit = -2
- $ f(2) = 3 $
- Since left and right limits are not equal, function is not continuous at $ x = 2 $

✘ Continuous: No

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#### Graph Sketching Tips:
- For $ x \leq 2 $: line $ y = 2x - 1 $, closed circle at $ x = 2 $
- For $ x > 2 $: line $ y = 2x - 6 $, open circle at $ x = 2 $
- Jump discontinuity at $ x = 2 $

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✔ Summary for Problem 3:

- Domain: Yes (all real numbers)
- Function: Yes
- Continuous: No
- $ f(-4) = -9 $
- $ f(0) = -1 $
- $ f(2) = 3 $

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✔ Final Answers Table:

| Problem | Domain? | Function? | Continuous? | $ f(-4) $ | $ f(0) $ | $ f(2) $ or $ f(3) $ |
|--------|---------|-----------|-------------|------------|------------|--------------------------|
| 1 | Yes | Yes | No | -12 | -3 | -3 |
| 2 | No | Yes | No | -5 | -1 | undefined |
| 3 | Yes | Yes | No | -9 | -1 | 3 |

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📌 Instructions for Graphs (on grid):

For each problem, draw:
- Two parts of the graph based on the piecewise rules
- Use open circles where the endpoint is excluded
- Use closed circles where included
- Show the break or jump at the boundary point

Let me know if you'd like me to generate a visual sketch or guide how to plot them step-by-step on the grid.