Piecewise function worksheet: Fill out & sign online | DocHub - Free Printable

Let's solve each of the piecewise functions and explain how to graph them step by step. Since I can't see the image directly, I’ll go through each problem based on the text you've provided.

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1.

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

#### Step-by-step:

- For $ x \leq 2 $: Graph $ f(x) = -x $. This is a line with slope $-1$, passing through the origin.
- At $ x = 2 $, $ f(2) = -2 $. Use a closed circle at $ (2, -2) $ because $ x \leq 2 $ includes 2.
- Extend this line to the left (for all $ x \leq 2 $).

- For $ x > 2 $: Graph $ f(x) = x $. This is a line with slope $1$.
- At $ x = 2 $, $ f(x) = x $ is not defined here (since it’s only for $ x > 2 $), so use an open circle at $ (2, 2) $.
- Draw the line $ y = x $ starting just right of $ x = 2 $, going upward.

#### Graph:
- Left side: Line $ y = -x $ from $ (-\infty, 2] $, ending at $ (2, -2) $ with a closed dot.
- Right side: Line $ y = x $ from $ (2, \infty) $, starting at $ (2, 2) $ with an open dot.

> Note: The two parts meet at different y-values at $ x=2 $, so there’s a "jump" in the graph.

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2.

$$
f(x) =
\begin{cases}
2 & \text{if } x > -3 \\
-5 & \text{if } x < -3
\end{cases}
$$

Wait — this function is undefined at $ x = -3 $ since both conditions are strict inequalities.

#### Step-by-step:

- For $ x > -3 $: $ f(x) = 2 $. A horizontal line at $ y = 2 $, but not including $ x = -3 $ → open circle at $ (-3, 2) $.
- For $ x < -3 $: $ f(x) = -5 $. A horizontal line at $ y = -5 $, but not including $ x = -3 $ → open circle at $ (-3, -5) $.

#### Graph:
- Horizontal line at $ y = 2 $ for $ x > -3 $, open circle at $ (-3, 2) $
- Horizontal line at $ y = -5 $ for $ x < -3 $, open circle at $ (-3, -5) $
- No point defined at $ x = -3 $

> This is a step function with a jump discontinuity at $ x = -3 $.

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3.

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

#### Step-by-step:

- For $ x \leq -2 $: $ f(x) = -1 $. Horizontal line at $ y = -1 $, including $ x = -2 $ → closed circle at $ (-2, -1) $
- For $ x > -2 $: $ f(x) = 2 $. Horizontal line at $ y = 2 $, excluding $ x = -2 $ → open circle at $ (-2, 2) $

#### Graph:
- Left: $ y = -1 $ from $ (-\infty, -2] $, closed dot at $ (-2, -1) $
- Right: $ y = 2 $ from $ (-2, \infty) $, open dot at $ (-2, 2) $

> Jump discontinuity at $ x = -2 $

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4.

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

#### Step-by-step:

- For $ x \leq -1 $: $ f(x) = -1 $ → horizontal line at $ y = -1 $, closed circle at $ (-1, -1) $
- For $ -1 < x < 1 $: $ f(x) = 1 $ → horizontal line at $ y = 1 $, open circles at both ends:
- Open at $ (-1, 1) $ (since $ x > -1 $)
- Open at $ (1, 1) $ (since $ x < 1 $)
- For $ x > 1 $: $ f(x) = x $ → line $ y = x $, starting after $ x = 1 $, open circle at $ (1, 1) $

#### Graph:
- Left: $ y = -1 $ for $ x \leq -1 $, closed dot at $ (-1, -1) $
- Middle: $ y = 1 $ from $ (-1, 1) $, open dots at $ (-1,1) $ and $ (1,1) $
- Right: $ y = x $ for $ x > 1 $, open dot at $ (1,1) $

> Note: At $ x = 1 $, $ f(x) = x $ starts at $ (1,1) $, but since $ x > 1 $, it's open. So no value at $ x=1 $ from this piece. But the middle piece also doesn’t include $ x=1 $. So $ f(1) $ is undefined? Wait — let's check:

Actually, the middle condition is $ -1 < x < 1 $, so $ x = 1 $ is not included. The third piece is $ x > 1 $, so $ x = 1 $ is not included either.

So $ f(1) $ is undefined. That means:
- At $ x = 1 $, no point exists.
- There is a gap at $ x = 1 $

But wait — let's double-check: Is there any case that includes $ x = 1 $? No. So function is undefined at $ x = 1 $.

However, if you intended $ x \geq 1 $, it might be different. But as written, it's not defined at $ x = 1 $.

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5.

$$
f(x) =
\begin{cases}
-x + 2 & \text{if } x \leq 0 \\
\frac{1}{2}x + 3 & \text{if } x > 0
\end{cases}
$$

#### Step-by-step:

- For $ x \leq 0 $: $ f(x) = -x + 2 $
- This is a line with slope $-1$, y-intercept $2$
- At $ x = 0 $, $ f(0) = -0 + 2 = 2 $ → closed circle at $ (0, 2) $
- Draw line to the left (negative $ x $)

- For $ x > 0 $: $ f(x) = \frac{1}{2}x + 3 $
- Slope $ \frac{1}{2} $, y-intercept $ 3 $
- At $ x = 0 $, $ f(0) = 3 $, but $ x > 0 $, so open circle at $ (0, 3) $
- Draw line to the right (positive $ x $)

#### Graph:
- Left part: Line $ y = -x + 2 $ from $ (-\infty, 0] $, closed dot at $ (0, 2) $
- Right part: Line $ y = \frac{1}{2}x + 3 $ from $ (0, \infty) $, open dot at $ (0, 3) $

> There is a jump discontinuity at $ x = 0 $, from $ y = 2 $ to $ y = 3 $

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6.

$$
f(x) =
\begin{cases}
x + 2 & \text{if } x \leq 2 \\
-\frac{1}{2}x + 4 & \text{if } x > 2
\end{cases}
$$

#### Step-by-step:

- For $ x \leq 2 $: $ f(x) = x + 2 $
- Line with slope $1$, y-intercept $2$
- At $ x = 2 $, $ f(2) = 2 + 2 = 4 $ → closed circle at $ (2, 4) $
- Draw line to the left

- For $ x > 2 $: $ f(x) = -\frac{1}{2}x + 4 $
- Slope $ -\frac{1}{2} $, y-intercept $ 4 $
- At $ x = 2 $, $ f(2) = -\frac{1}{2}(2) + 4 = -1 + 4 = 3 $
- But since $ x > 2 $, we don’t include $ x = 2 $ → open circle at $ (2, 3) $
- Draw line to the right

#### Graph:
- Left: $ y = x + 2 $ for $ x \leq 2 $, closed dot at $ (2, 4) $
- Right: $ y = -\frac{1}{2}x + 4 $ for $ x > 2 $, open dot at $ (2, 3) $

> Jump discontinuity at $ x = 2 $: from $ y = 4 $ to $ y = 3 $, but the second piece starts at $ y = 3 $, so there's a gap between $ (2,3) $ and $ (2,4) $

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✔ Summary of Key Steps for Graphing Piecewise Functions:

1. Identify domains for each piece.
2. Determine if endpoints are included (use closed or open circles).
3. Graph each piece separately within its domain.
4. Check continuity at boundary points.
5. Label key points and ensure correct connections.

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📌 Final Tips:

- Always check the inequality signs: $ \leq $ or $ \geq $ → closed circle; $ < $ or $ > $ → open circle.
- When switching pieces, make sure you're not missing any values.
- If a value is not covered by any piece (like in #4 at $ x = 1 $), then the function is undefined there.

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If you'd like, I can generate visual descriptions or ASCII representations of the graphs for each one. Let me know!