Piecewise function graph showing a line and a horizontal segment with a 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 graph it for $ x < -2 $
- At $ x = -2 $, we do not include this point (since it's strictly less than), so we use an open circle at $ x = -2 $

Let’s 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: $ (-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 in this piece.

Wait! Let's check what happens at $ x = -2 $:

Since $ x \geq -2 $ uses the second rule: $ f(x) = -4 $, so $ f(-2) = -4 $

But from the first rule, as $ x \to -2^- $, $ f(x) \to -2 + 5 = 3 $

So the left-hand limit is 3, but the function value is -4 → discontinuity!

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

- This is a horizontal line at $ y = -4 $
- Start at $ x = -2 $, include that point (closed circle), and extend to the right

Point: $ (-2, -4) $, then $ (0, -4), (2, -4) $, etc.

#### Step 3: Graph

- Draw a line for $ x + 5 $ from the left up to $ x = -2 $, open circle at $ (-2, 3) $
- Draw a horizontal line at $ y = -4 $ starting at $ (-2, -4) $ with a closed circle, extending to the right

#### Step 4: Check for continuity at $ x = -2 $

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

Since left-hand limit ≠ right-hand limit, and also ≠ function value (but even if they matched, the jump would still be there), this is a jump discontinuity at $ x = -2 $

✔ Discontinuity at $ x = -2 $

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Graph 1 Summary:

- Line $ y = x + 5 $ for $ x < -2 $, ending with open circle at $ (-2, 3) $
- Horizontal line $ y = -4 $ for $ x \geq -2 $, starting with closed circle at $ (-2, -4) $
- Jump discontinuity 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 $ — linear, slope 2
- For $ x \geq 1 $: $ f(x) = -2x + 3 $ — linear, slope -2

#### 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 $
- Plot these and draw a ray going to the left from $ x = 1 $
- At $ x = 1 $, do not include this point (since $ x < 1 $), so open circle at $ (1, 2x+1) = (1, 3) $

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

- Find some points:
- $ x = 1 $: $ f(1) = -2(1) + 3 = 1 $
- $ x = 2 $: $ f(2) = -4 + 3 = -1 $
- $ x = 3 $: $ f(3) = -6 + 3 = -3 $
- Start at $ (1,1) $, closed circle, since $ x \geq 1 $
- Draw line to the right

#### 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-hand limit = 3
- Right-hand limit = 1
- $ f(1) = 1 $

Since left-hand limit ≠ right-hand limit → jump discontinuity at $ x = 1 $

Even though the right side matches the function value, the left side doesn't approach the same value → discontinuous at $ x = 1 $

✔ Discontinuity at $ x = 1 $

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Graph 2 Summary:

- Line $ y = 2x + 1 $ for $ x < 1 $, ending with open circle at (1, 3)
- Line $ y = -2x + 3 $ for $ x \geq 1 $, starting with closed circle at (1, 1), going down
- Jump discontinuity at $ x = 1 $

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✔ Final Answers:

#### 1.
- Graph:
- Ray of $ y = x + 5 $ for $ x < -2 $, open circle at $ (-2, 3) $
- Horizontal line $ y = -4 $ for $ x \geq -2 $, closed circle at $ (-2, -4) $
- Discontinuity at $ x = -2 $ (jump)

#### 2.
- Graph:
- Ray of $ y = 2x + 1 $ for $ x < 1 $, open circle at $ (1, 3) $
- Ray of $ y = -2x + 3 $ for $ x \geq 1 $, closed circle at $ (1, 1) $
- Discontinuity at $ x = 1 $ (jump)

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Let me know if you'd like me to sketch the graphs digitally or provide coordinates for plotting!