Solved Math 2 Name: Piecewise Functions Worksheet #2 Part I ... - Free Printable

Let's solve each piecewise function step by step, graph them, and identify any points of discontinuity.

---

Problem 1:

$$
f(x) =
\begin{cases}
x + 5 & \text{if } x < -2 \\
-4 & \text{if } x \geq -2
\end{cases}
$$

#### Step 1: Analyze the pieces

- For $ x < -2 $: This is a linear function $ f(x) = x + 5 $
- For $ x \geq -2 $: This is a constant function $ f(x) = -4 $

#### Step 2: Graph each part

Part A: $ f(x) = x + 5 $ for $ x < -2 $

- This is a straight line with slope 1 and y-intercept 5.
- But we only draw it for $ x < -2 $
- At $ x = -2 $, we do not include this point (since it's strictly less than), so use an open circle at $ x = -2 $

Find a few points:
- When $ x = -3 $: $ f(-3) = -3 + 5 = 2 $
- When $ x = -4 $: $ f(-4) = -4 + 5 = 1 $
- When $ x = -5 $: $ f(-5) = 0 $

So plot points like (-5, 0), (-4, 1), (-3, 2), and draw a ray going to the left from $ x = -2 $, with an open circle at $ (-2, 3) $ because:
- $ f(-2) $ is not defined by this rule, but let’s check the next piece.

Wait — at $ x = -2 $, since $ x \geq -2 $, we use the second rule.

Part B: $ f(x) = -4 $ for $ x \geq -2 $

- This is a horizontal line at $ y = -4 $
- Include $ x = -2 $, so closed circle at $ (-2, -4) $
- Draw this line to the right from $ x = -2 $ onward

Now compare:

- Left side: as $ x \to -2^- $, $ f(x) = x + 5 \to -2 + 5 = 3 $
- Right side: $ f(-2) = -4 $

So there is a jump from $ y = 3 $ (approaching from the left) to $ y = -4 $ at $ x = -2 $

#### ✔ Discontinuity at $ x = -2 $ — Jump discontinuity

---

📈 Graph Description (for Problem 1):

- Draw a line segment for $ f(x) = x+5 $ for $ x < -2 $: goes through points like (-5,0), (-4,1), (-3,2), ending with an open circle at (-2, 3).
- Then draw a horizontal line $ y = -4 $ starting at $ x = -2 $: closed circle at (-2, -4), and extend to the right.

👉 Point of discontinuity: $ x = -2 $

---

Problem 2:

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

#### Step 1: Analyze the pieces

- For $ x < 1 $: $ f(x) = 2x + 1 $ → line with slope 2, y-intercept 1
- For $ x \geq 1 $: $ f(x) = -2x + 3 $ → line with slope -2, y-intercept 3

#### Step 2: Graph each part

Part A: $ f(x) = 2x + 1 $ for $ x < 1 $

- Find some points:
- $ x = 0 $: $ f(0) = 1 $
- $ x = -1 $: $ f(-1) = -2 + 1 = -1 $
- $ x = 0.5 $: $ f(0.5) = 1 + 1 = 2 $
- As $ x \to 1^- $: $ f(x) \to 2(1) + 1 = 3 $
- So approach $ (1, 3) $, but open circle at $ (1, 3) $ since $ x < 1 $

Part B: $ f(x) = -2x + 3 $ for $ x \geq 1 $

- At $ x = 1 $: $ f(1) = -2(1) + 3 = 1 $
- So closed circle at (1, 1)
- Other points:
- $ x = 2 $: $ f(2) = -4 + 3 = -1 $
- $ x = 3 $: $ f(3) = -6 + 3 = -3 $
- Draw line from $ (1,1) $ going down to the right

Now check continuity at $ x = 1 $:

- Left-hand limit: $ \lim_{x \to 1^-} f(x) = 2(1) + 1 = 3 $
- Right-hand limit: $ \lim_{x \to 1^+} f(x) = -2(1) + 3 = 1 $
- Function value: $ f(1) = 1 $

Since left-hand limit (3) ≠ right-hand limit (1), and also ≠ function value, there is a jump.

#### ✔ Discontinuity at $ x = 1 $ — Jump discontinuity

---

📈 Graph Description (for Problem 2):

- For $ x < 1 $: draw line $ y = 2x + 1 $, with open circle at $ (1, 3) $
- For $ x \geq 1 $: draw line $ y = -2x + 3 $, with closed circle at $ (1, 1) $
- The two parts do not meet, so there's a jump from $ y = 3 $ to $ y = 1 $ at $ x = 1 $

---

✔ Final Answers:

#### Problem 1:
- Graph: Line $ y = x+5 $ for $ x < -2 $ (open circle at $ (-2,3) $), then horizontal line $ y = -4 $ for $ x \geq -2 $ (closed circle at $ (-2,-4) $)
- Discontinuity at $ x = -2 $ → Jump discontinuity

#### Problem 2:
- Graph: Line $ y = 2x+1 $ for $ x < 1 $ (open circle at $ (1,3) $), then line $ y = -2x+3 $ for $ x \geq 1 $ (closed circle at $ (1,1) $)
- Discontinuity at $ x = 1 $ → Jump discontinuity

---

📝 Summary:

| Problem | Discontinuity? | Type | Location |
|--------|----------------|------|----------|
| 1 | Yes | Jump | $ x = -2 $ |
| 2 | Yes | Jump | $ x = 1 $ |

> 🔍 Note: Both functions are continuous everywhere except at the boundary point where the definition changes. Since the left and right limits don’t match, they have jump discontinuities.

Let me know if you'd like a visual sketch or help drawing it on paper!