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, and evaluate specific values.

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Problem 1:

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

#### Step 1: Domain
- The function is defined for all real numbers:
- $ x < -2 $: defined by $ x - 8 $
- $ x \geq -2 $: defined by $ x^2 - 2x - 3 $
- So, domain: All real numbers → Yes, domain is all real numbers.

#### Step 2: Continuity at $ x = -2 $
We check if the left-hand limit (LHL), right-hand limit (RHL), and function value at $ x = -2 $ are equal.

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

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

- Function value at $ x = -2 $:
Since $ x = -2 \geq -2 $, use second case:
$$
f(-2) = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5
$$

Since LHL = -10 ≠ RHL = 5, the function has a jump discontinuity at $ x = -2 $.
So, not continuous at $ x = -2 $, but it's continuous elsewhere.

→ Continuous? No

#### Step 3: Evaluate Function Values

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

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

- $ f(2) $: $ 2 \geq -2 $, so use same formula
$$
f(2) = (2)^2 - 2(2) - 3 = 4 - 4 - 3 = -3
$$

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✔ Summary for Problem 1:

| Question | Answer |
|--------|--------|
| Domain: Yes/No | Yes |
| Continuous: Yes/No | No |
| $ f(-4) $ | -12 |
| $ f(0) $ | -3 |
| $ f(2) $ | -3 |

---

Problem 2:

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

Note: There is no definition at $ x = 3 $.

#### Step 1: Domain
- Defined for $ x < 3 $ and $ x > 3 $, but not at $ x = 3 $.
- So, domain is $ (-\infty, 3) \cup (3, \infty) $
→ Domain: No (not all real numbers)

#### Step 2: Continuity
- Since $ f(x) $ is not defined at $ x = 3 $, it cannot be continuous there.
- But we can check limits to see if it could be made continuous.

- Left-hand limit ($ x \to 3^- $): use $ 2x - 1 $
$$
\lim_{x \to 3^-} f(x) = 2(3) - 1 = 6 - 1 = 5
$$

- Right-hand limit ($ x \to 3^+ $): use $ x^2 - 3 $
$$
\lim_{x \to 3^+} f(x) = (3)^2 - 3 = 9 - 3 = 6
$$

- LHL = 5, RHL = 6 → Not equal, and no value at $ x = 3 $

→ Not continuous (and undefined at $ x = 3 $)

→ Continuous? No

#### Step 3: Evaluate Function Values

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

- $ f(0) $: $ 0 < 3 $, so use $ 2x - 1 $
$$
f(0) = 2(0) - 1 = -1
$$

- $ f(5) $: $ 5 > 3 $, so use $ x^2 - 3 $
$$
f(5) = (5)^2 - 3 = 25 - 3 = 22
$$

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✔ Summary for Problem 2:

| Question | Answer |
|--------|--------|
| Domain: Yes/No | No |
| Continuous: Yes/No | No |
| $ f(-2) $ | -5 |
| $ f(0) $ | -1 |
| $ f(5) $ | 22 |

---

Problem 3:

Function:
$$
f(x) =
\begin{cases}
2x - 1 & \text{if } x \leq 2 \\
3x - 6 & \text{if } x > 2
\end{cases}
$$

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

#### Step 2: Continuity at $ x = 2 $
Check if left-hand limit, right-hand limit, and function value match.

- Left-hand limit ($ x \to 2^- $): use $ 2x - 1 $
$$
\lim_{x \to 2^-} f(x) = 2(2) - 1 = 4 - 1 = 3
$$

- Right-hand limit ($ x \to 2^+ $): use $ 3x - 6 $
$$
\lim_{x \to 2^+} f(x) = 3(2) - 6 = 6 - 6 = 0
$$

- Function value at $ x = 2 $: since $ x \leq 2 $, use $ 2x - 1 $
$$
f(2) = 2(2) - 1 = 3
$$

LHL = 3, RHL = 0 → Not equal, so discontinuous at $ x = 2 $

→ Continuous? No

#### Step 3: Evaluate Function Values

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

- $ f(0) $: $ 0 \leq 2 $, so use $ 2x - 1 $
$$
f(0) = 2(0) - 1 = -1
$$

- $ f(3) $: $ 3 > 2 $, so use $ 3x - 6 $
$$
f(3) = 3(3) - 6 = 9 - 6 = 3
$$

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✔ Summary for Problem 3:

| Question | Answer |
|--------|--------|
| Domain: Yes/No | Yes |
| Continuous: Yes/No | No |
| $ f(-4) $ | -9 |
| $ f(0) $ | -1 |
| $ f(3) $ | 3 |

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📌 Final Answers:

#### 1.
- Domain: Yes
- Continuous: No
- $ f(-4) = -12 $
- $ f(0) = -3 $
- $ f(2) = -3 $

#### 2.
- Domain: No
- Continuous: No
- $ f(-2) = -5 $
- $ f(0) = -1 $
- $ f(5) = 22 $

#### 3.
- Domain: Yes
- Continuous: No
- $ f(-4) = -9 $
- $ f(0) = -1 $
- $ f(3) = 3 $

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🎯 Graphing Tips (for sketching):

For each function, plot both pieces on the coordinate plane:
- Use open circles where a piece is not included (e.g., $ x < 3 $, so open circle at $ x = 3 $)
- Use closed circles where a piece includes the point (e.g., $ x \leq 2 $, so closed at $ x = 2 $)
- Draw the lines or curves accordingly.

Let me know if you'd like help sketching any of them!