Graph of a piecewise function with a jump discontinuity at x = -2.

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

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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 $, since it's not included (strict inequality), we 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 $
- So plot points like $ (-4,1), (-3,2) $, and draw a ray going to the left, ending with an open circle at $ x = -2 $

At $ x = -2 $:
$ f(-2) = -4 $ from the second rule, so we’ll have a solid dot at $ (-2, -4) $

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

- This is a horizontal line at $ y = -4 $
- Starts at $ x = -2 $ (inclusive), so solid dot at $ (-2, -4) $
- Continue to the right indefinitely

#### Step 3: Check for continuity at $ x = -2 $

We need to check if the left-hand limit, right-hand limit, and function value agree.

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

Since:
- $ \lim_{x \to -2^-} f(x) = 3 $
- $ \lim_{x \to -2^+} f(x) = -4 $
- $ f(-2) = -4 $

The left and right limits are not equal, so there is a jump discontinuity at $ x = -2 $

✔ Discontinuity at $ x = -2 $

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✔ Graph Description for Problem 1:

- Draw a line $ y = x + 5 $ for $ x < -2 $, ending with an open circle at $ (-2, 3) $
- Draw a horizontal line $ y = -4 $ starting at $ (-2, -4) $ with a solid dot
- There is a jump from $ y = 3 $ to $ y = -4 $ at $ x = -2 $

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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 points:
- $ x = 0 $: $ f(0) = 1 $
- $ x = -1 $: $ f(-1) = -2 + 1 = -1 $
- So plot $ (-1,-1), (0,1) $
- Draw a line going to the left, but stop before $ x = 1 $
- At $ x = 1 $, since it's not included, use an open circle at $ x = 1 $

Value at $ x = 1 $ from this side: $ 2(1) + 1 = 3 $ → so open circle at $ (1, 3) $

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

- Find values:
- $ x = 1 $: $ f(1) = -2(1) + 3 = 1 $
- $ x = 2 $: $ f(2) = -4 + 3 = -1 $
- $ x = 3 $: $ f(3) = -6 + 3 = -3 $
- So plot $ (1,1), (2,-1), (3,-3) $
- Since $ x \geq 1 $, start with a solid dot at $ (1,1) $

#### Step 3: 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) = -2(1) + 3 = 1 $

So:
- Left limit = 3
- Right limit = 1
- $ f(1) = 1 $

Since left and right limits are not equal, there is a jump discontinuity at $ x = 1 $

✔ Discontinuity at $ x = 1 $

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✔ Graph Description for Problem 2:

- Draw $ y = 2x + 1 $ for $ x < 1 $, ending with an open circle at $ (1, 3) $
- Draw $ y = -2x + 3 $ for $ x \geq 1 $, starting with a solid dot at $ (1, 1) $
- There is a jump from $ y = 3 $ to $ y = 1 $ at $ x = 1 $

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✔ 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 $ (solid dot at $ (-2,-4) $)
- Discontinuity at $ x = -2 $ (jump)

#### Problem 2:
- Graph: Line $ y = 2x + 1 $ for $ x < 1 $ (open circle at $ (1,3) $), then line $ y = -2x + 3 $ for $ x \geq 1 $ (solid dot at $ (1,1) $)
- Discontinuity at $ x = 1 $ (jump)

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📌 Summary of Discontinuities:

| Problem | Discontinuity at | Type |
|--------|------------------|------|
| 1 | $ x = -2 $ | Jump |
| 2 | $ x = 1 $ | Jump |

> 🔍 Both functions have jump discontinuities at the boundary point where the definition changes.

Let me know if you'd like me to generate the actual graphs or help sketch them on paper!