Piecewise Functions Practice Worksheet with Handwritten Solutions

Let's go through the image and solve each part step by step, verifying your work and explaining the reasoning behind each solution.

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Part 1: Evaluate Piecewise Functions at Given Points

We are given three piecewise functions:

#### **1. $ f(x) = \begin{cases}
x^3 & \text{if } x < 0 \\
-4 & \text{if } x = 0 \\
x^2 + 5 & \text{if } x > 0
\end{cases} $**

Evaluate:
- a) $ f(-1) $
Since $ -1 < 0 $, use $ f(x) = x^3 $:
$ f(-1) = (-1)^3 = -1 $ ✔

- b) $ f(0) $
$ x = 0 $ → use $ f(x) = -4 $:
$ f(0) = -4 $ ✔

- c) $ f(1) $
$ 1 > 0 $ → use $ f(x) = x^2 + 5 $:
$ f(1) = 1^2 + 5 = 6 $ ✔

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#### **2. $ g(x) = \begin{cases}
x + 2 & \text{if } x \geq 3 \\
-x + 2 & \text{if } x < 3
\end{cases} $**

- d) $ g(0) $
$ 0 < 3 $ → use $ -x + 2 $:
$ g(0) = -0 + 2 = 2 $ ✔

- e) $ g(3) $
$ 3 \geq 3 $ → use $ x + 2 $:
$ g(3) = 3 + 2 = 5 $ ✔

- f) $ g(6) $
$ 6 \geq 3 $ → use $ x + 2 $:
$ g(6) = 6 + 2 = 8 $ ✔

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#### **3. $ h(x) = \begin{cases}
|x| & \text{if } -3 < x < 4 \\
x - 1 & \text{if } x = 4 \\
2x^2 & \text{if } 4 < x < 6
\end{cases} $**

Note: The condition is not defined for $ x \leq -3 $, so domain starts from $ x > -3 $.

- g) $ h(-1) $
$ -3 < -1 < 4 $ → use $ |x| $:
$ h(-1) = |-1| = 1 $ ✔

- h) $ h(4) $
$ x = 4 $ → use $ x - 1 $:
$ h(4) = 4 - 1 = 3 $ ✔

- i) $ h(7) $
$ 7 > 6 $, and no definition for $ x \geq 6 $, so undefined → $ \emptyset $ ✔

✔ All evaluations are correct.

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Part 2: Graphing and Finding Domain & Range

Now we analyze the graphs of the piecewise functions and determine their domains and ranges.

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#### Function k(x):
$$
k(x) = \begin{cases}
x^2 & \text{if } 0 \leq x < 2 \\
2x - 4 & \text{if } x > 2
\end{cases}
$$

Graph:
- From $ x = 0 $ to $ x = 2 $: parabola $ y = x^2 $, closed at (0,0), open at (2,4)
- For $ x > 2 $: line $ y = 2x - 4 $, starting just after $ x = 2 $

Domain (D):
All $ x \geq 0 $, but no value at $ x = 2 $ in first piece, second piece starts at $ x > 2 $. So:
- $ [0, 2) \cup (2, \infty) $ → $ D: [0, 2) \cup (2, \infty) $ ✔

Range (R):
- $ x^2 $ on $ [0,2) $ → outputs $ [0, 4) $
- $ 2x - 4 $ for $ x > 2 $ → as $ x \to 2^+ $, $ y \to 0 $; as $ x \to \infty $, $ y \to \infty $
- So $ y > 0 $, but note: at $ x = 2 $, $ y = 0 $ is not included in this branch.
- But $ y = 0 $ is achieved at $ x = 0 $
- So overall range: $ [0, 4) \cup (0, \infty) = [0, \infty) $ ✔

✔ Your answer: $ D: [0, 2) \cup (2, \infty), R: [0, \infty) $ → Correct

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#### Function m(x):
$$
m(x) = \begin{cases}
-x^2 + 6 & \text{if } x < -2 \\
|x| & \text{if } -2 \leq x < 3 \\
2x - 5 & \text{if } x \geq 3
\end{cases}
$$

Graph Analysis:
- $ x < -2 $: downward parabola $ -x^2 + 6 $
- As $ x \to -\infty $, $ y \to -\infty $
- At $ x = -2^- $, $ y = -(-2)^2 + 6 = -4 + 6 = 2 $
- $ -2 \leq x < 3 $: $ |x| $ → V-shape
- From $ x = -2 $ to $ 0 $: $ -x $, then $ x $
- At $ x = -2 $: $ y = 2 $
- At $ x = 0 $: $ y = 0 $
- At $ x = 3^- $: $ y = 3 $
- $ x \geq 3 $: $ 2x - 5 $
- At $ x = 3 $: $ y = 6 - 5 = 1 $
- Increases to $ \infty $

Domain (D): All real numbers → $ (-\infty, \infty) $ ✔

Range (R):
- Left side ($ x < -2 $): $ -x^2 + 6 $, max at $ x = 0 $, but not in this region. At $ x = -2 $, $ y = 2 $, and goes down to $ -\infty $
- So $ (-\infty, 2] $
- Middle: $ |x| $ on $ [-2, 3) $
- $ y \in [0, 3) $ — since $ x = 3 $ not included
- Right: $ 2x - 5 $ for $ x \geq 3 $: starts at $ y = 1 $, increases to $ \infty $
- So $ [1, \infty) $

Now combine:
- From left: $ (-\infty, 2] $
- Middle: $ [0, 3) $
- Right: $ [1, \infty) $

Union: $ (-\infty, 2] \cup [0, 3) \cup [1, \infty) = (-\infty, \infty) $

Wait! Is there a gap?

No: $ (-\infty, 2] $ covers everything up to 2, and $ [1, \infty) $ covers from 1 onward. Together, they cover all real numbers.

But let’s check: can $ y = 0.5 $? Yes, from middle piece at $ x = 0.5 $. Can $ y = -1 $? Yes, from left piece: solve $ -x^2 + 6 = -1 $ → $ x^2 = 7 $ → $ x = \pm\sqrt{7} $, $ \sqrt{7} \approx 2.6 < 2 $? No, $ \sqrt{7} \approx 2.6 > 2 $, so $ x = -\sqrt{7} \approx -2.6 < -2 $, valid.

So yes, all values are covered.

✔ Your answer: $ D: (-\infty, \infty), R: (-\infty, \infty) $ → Correct

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#### Function p(x):
$$
p(x) = \begin{cases}
x + 3 & \text{if } -3 \leq x < 0 \\
-2 & \text{if } 0 \leq x < 4 \\
-2x + 6 & \text{if } x \geq 4
\end{cases}
$$

Graph:
- $ -3 \leq x < 0 $: line $ y = x + 3 $
- At $ x = -3 $: $ y = 0 $
- At $ x = 0^- $: $ y = 3 $
- $ 0 \leq x < 4 $: constant $ y = -2 $
- So horizontal line at $ y = -2 $
- $ x \geq 4 $: $ y = -2x + 6 $
- At $ x = 4 $: $ y = -8 + 6 = -2 $
- Then decreases to $ -\infty $

Domain (D): All $ x \geq -3 $ → $ [-3, \infty) $ ✔

Range (R):
- First piece: $ x + 3 $ on $ [-3, 0) $ → $ y \in [0, 3) $
- Second: $ y = -2 $ → $ \{-2\} $
- Third: $ -2x + 6 $, $ x \geq 4 $ → at $ x = 4 $: $ y = -2 $, then decreases → $ (-\infty, -2] $

Now combine:
- $ [0, 3) $
- $ \{-2\} $
- $ (-\infty, -2] $

Total: $ (-\infty, -2] \cup \{-2\} \cup [0, 3) = (-\infty, -2] \cup [0, 3) $

But $ \{-2\} $ is already in $ (-\infty, -2] $, so final range: $ (-\infty, -2] \cup [0, 3) $

Wait — you wrote: $ R: (-\infty, -2] \cup [0, 3) $ — but you have written $ R: (-\infty, -2] \cup [0, 3) $ — actually, you wrote $ R: (-\infty, -2] \cup [0, 3) $, which is correct!

Wait — in your handwriting, it says:
> $ D: [-3, \infty) $
> $ R: (-\infty, -2] \cup [0, 3) $

✔ Yes, that's correct.

But wait — what about $ y = -1 $? Is it in range? No. $ y = -1.5 $? No. Only $ y = -2 $ is in middle and third piece.

And $ y = 0 $ is in first piece (at $ x = -3 $), and also at $ x = 0 $ in second piece? No — at $ x = 0 $, $ y = -2 $, so $ y = 0 $ only from first piece.

So range:
- $ (-\infty, -2] $: from third piece
- $ \{-2\} $: from second
- $ [0, 3) $: from first

So total: $ (-\infty, -2] \cup [0, 3) $

✔ Correct.

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Part 3: Write a Piecewise Function from Graphs

Now we interpret the graphs and write equations.

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#### First graph (left):
- Line from bottom-left to open circle at $ (0,1) $: slope?
- Let’s assume it goes through $ (-2,-1) $, $ (-1,0) $, $ (0,1) $ → open at $ (0,1) $
- Slope: $ \frac{1 - 0}{0 - (-1)} = 1 $
- Equation: $ y = x + 1 $ for $ x < 0 $
- Then for $ x > 0 $: curve going up, looks like $ y = x^2 $
- At $ x = 1 $, $ y = 1 $; $ x = 2 $, $ y = 4 $ → yes
- Closed at $ (0,0) $? Wait — at $ x = 0 $, is there a point?

Looking at graph:
- Open circle at $ (0,1) $ → so not included
- Closed point at $ (0,0) $? No — seems like the right piece starts at $ (0,0) $ with a closed dot?

Wait — your function:
> $ \text{becky}(x) = \begin{cases} 3x + 1 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases} $

But let’s test:
- At $ x = -1 $: $ 3(-1) + 1 = -2 $ → but graph shows $ y = 0 $ at $ x = -1 $
- So slope must be 1, not 3

Wait — maybe you misread the graph?

Let’s re-analyze:
- From $ x = -2 $ to $ x = 0 $: passes through $ (-2,-1) $, $ (-1,0) $, $ (0,1) $ → open circle at $ (0,1) $
- So equation: $ y = x + 1 $
- For $ x > 0 $: curve $ y = x^2 $, but at $ x = 0 $, $ y = 0 $, so closed dot at (0,0)

So function should be:
$$
f(x) = \begin{cases}
x + 1 & \text{if } x < 0 \\
x^2 & \text{if } x \geq 0
\end{cases}
$$

But you wrote:
> $ \text{becky}(x) = \begin{cases} 3x + 1 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases} $

This is incorrect because at $ x = -1 $, $ 3(-1)+1 = -2 $, but graph shows $ y = 0 $

So your answer is wrong here.

✔ Correction: $ f(x) = \begin{cases} x + 1 & x < 0 \\ x^2 & x \geq 0 \end{cases} $

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#### Second graph (middle):
Three parts:
- Left: line from $ (-3, -1) $ to $ (-1, 1) $, with open circle at $ (-1,1) $
- Middle: V-shaped, from $ (-1,1) $ to $ (0,0) $, then down to $ (1,-1) $, open at $ (1,-1) $
- Right: line from $ (1,-1) $ to $ (2,-2) $, closed at both ends

Wait — your function:
> $ f(x) = \begin{cases}
\frac{4}{3}(x+3) + 1 & \text{if } -6 \leq x \leq -3 \\
2 & \text{if } -3 < x < -1 \\
-|x| & \text{if } -1 \leq x \leq 1
\end{cases} $

Wait — this doesn't match.

Let’s analyze properly.

From graph:
- Left segment: from $ (-3, -1) $ to $ (-1, 1) $, open at $ (-1,1) $
- Slope: $ \frac{1 - (-1)}{-1 - (-3)} = \frac{2}{2} = 1 $
- Point-slope: $ y + 1 = 1(x + 3) $ → $ y = x + 2 $
- But at $ x = -3 $: $ y = -1 $, yes; $ x = -1 $: $ y = 1 $, but open circle → so $ x < -1 $
- So: $ y = x + 2 $ for $ -3 \leq x < -1 $
- Middle: from $ (-1,1) $ to $ (0,0) $, then to $ (1,-1) $, symmetric
- This is $ y = -|x| $ for $ -1 \leq x \leq 1 $
- At $ x = 0 $: $ y = 0 $
- At $ x = 1 $: $ y = -1 $
- At $ x = -1 $: $ y = -1 $? But graph has $ y = 1 $ at $ x = -1 $
- Wait — contradiction

Wait — at $ x = -1 $, the left segment ends at $ y = 1 $, but middle starts at $ y = ? $

Actually, at $ x = -1 $, the graph has a closed dot at (-1,1)? Or open?

Looking at your graph: it appears that the V-shape starts at $ (-1,1) $, goes down to $ (0,0) $, then to $ (1,-1) $, and open at (1,-1)

But $ y = -|x| $ at $ x = -1 $ is $ -1 $, not 1.

So not $ -|x| $

Wait — perhaps it's $ y = -|x + 1| + 1 $? Let's try.

At $ x = -1 $: $ y = -|0| + 1 = 1 $ → good
At $ x = 0 $: $ y = -|1| + 1 = 0 $ → good
At $ x = 1 $: $ y = -|2| + 1 = -1 $ → good

So: $ y = -|x + 1| + 1 $ for $ -1 \leq x \leq 1 $

Then right segment: from $ (1,-1) $ to $ (2,-2) $, closed at both ends
- Slope: $ \frac{-2 - (-1)}{2 - 1} = -1 $
- Equation: $ y + 1 = -1(x - 1) $ → $ y = -x $
- At $ x = 1 $: $ y = -1 $, at $ x = 2 $: $ y = -2 $

So piecewise:
$$
f(x) = \begin{cases}
x + 2 & \text{if } -3 \leq x < -1 \\
-|x + 1| + 1 & \text{if } -1 \leq x \leq 1 \\
-x & \text{if } 1 < x \leq 2
\end{cases}
$$

But you wrote:
> $ f(x) = \begin{cases}
\frac{4}{3}(x+3) + 1 & \text{if } -6 \leq x \leq -3 \\
2 & \text{if } -3 < x < -1 \\
-|x| & \text{if } -1 \leq x \leq 1
\end{cases} $

This is completely off.

- $ \frac{4}{3}(x+3) + 1 $: at $ x = -3 $: $ 0 + 1 = 1 $, but graph has $ y = -1 $
- And "2" for $ -3 < x < -1 $? That would be horizontal line at $ y = 2 $, but graph shows line from $ (-3,-1) $ to $ (-1,1) $

So your answer is incorrect

✔ Correct function:
$$
f(x) = \begin{cases}
x + 2 & \text{if } -3 \leq x < -1 \\
-|x + 1| + 1 & \text{if } -1 \leq x \leq 1 \\
-x & \text{if } 1 < x \leq 2
\end{cases}
$$

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#### Third graph (right):
- Line from $ (-3, 2) $ to $ (-1, 4) $, closed at both ends
- Then from $ (-1, 4) $ to $ (1, -2) $, open at $ (-1,4) $, closed at $ (1,-2) $
- Then from $ (1,-2) $ to $ (2,-1) $, open at $ (1,-2) $, closed at $ (2,-1) $

Wait — your function:
> $ my(x) = \begin{cases}
\frac{1}{2}(x+1) + 5 & \text{if } x < -1 \\
-2x - 2 & \text{if } -1 < x < 2 \\
\frac{1}{3}(x+2) - 6 & \text{if } x \geq 2
\end{cases} $

Check:

First piece: $ x < -1 $
- $ \frac{1}{2}(x+1) + 5 $
- At $ x = -3 $: $ \frac{1}{2}(-2) + 5 = -1 + 5 = 4 $, but graph shows $ y = 2 $ → too high

Try slope:
From $ (-3,2) $ to $ (-1,4) $: slope = $ \frac{4-2}{-1+3} = \frac{2}{2} = 1 $
Equation: $ y - 2 = 1(x + 3) $ → $ y = x + 5 $
At $ x = -3 $: $ y = 2 $, $ x = -1 $: $ y = 4 $ → correct

So: $ y = x + 5 $ for $ x < -1 $

Middle: $ -1 < x < 2 $
From $ (-1,4) $ to $ (1,-2) $: slope = $ \frac{-2 - 4}{1 + 1} = \frac{-6}{2} = -3 $
Equation: $ y - 4 = -3(x + 1) $ → $ y = -3x + 1 $
At $ x = 1 $: $ y = -3 + 1 = -2 $ → good

But you wrote $ -2x - 2 $: at $ x = 1 $: $ -2 - 2 = -4 $ → wrong

Right: $ x \geq 2 $
From $ (2,-1) $: if we extend line from $ (1,-2) $ to $ (2,-1) $, slope = $ \frac{-1 + 2}{2 - 1} = 1 $
Equation: $ y + 2 = 1(x - 1) $ → $ y = x - 3 $
At $ x = 2 $: $ y = -1 $ → good

But you wrote $ \frac{1}{3}(x+2) - 6 $: at $ x = 2 $: $ \frac{1}{3}(4) - 6 = \frac{4}{3} - 6 = -\frac{14}{3} \ne -1 $

So your function is incorrect

✔ Correct function:
$$
my(x) = \begin{cases}
x + 5 & \text{if } x < -1 \\
-3x + 1 & \text{if } -1 < x < 2 \\
x - 3 & \text{if } x \geq 2
\end{cases}
$$

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✔ Final Summary

#### Corrected Answers:

1. Evaluations: All correct ✔

2. Domains and Ranges:
- $ k(x) $: $ D: [0, 2) \cup (2, \infty), R: [0, \infty) $ ✔
- $ m(x) $: $ D: (-\infty, \infty), R: (-\infty, \infty) $ ✔
- $ p(x) $: $ D: [-3, \infty), R: (-\infty, -2] \cup [0, 3) $ ✔

3. Piecewise Functions from Graphs:
- First graph:
$$
f(x) = \begin{cases}
x + 1 & \text{if } x < 0 \\
x^2 & \text{if } x \geq 0
\end{cases}
$$
- Second graph:
$$
f(x) = \begin{cases}
x + 2 & \text{if } -3 \leq x < -1 \\
-|x + 1| + 1 & \text{if } -1 \leq x \leq 1 \\
-x & \text{if } 1 < x \leq 2
\end{cases}
$$
- Third graph:
$$
my(x) = \begin{cases}
x + 5 & \text{if } x < -1 \\
-3x + 1 & \text{if } -1 < x < 2 \\
x - 3 & \text{if } x \geq 2
\end{cases}
$$

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Let me know if you'd like these graphs sketched or explained further!