Examples of piecewise-defined functions with graphs, equations, and evaluation tasks.

Let's solve each part of the given problems step by step, focusing on piecewise-defined functions.

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Example 3:

Given:
$$
f(x) =
\begin{cases}
-\frac{3}{2}x + 6, & x > 2 \\
2x - 1, & -3 < x \leq 2
\end{cases}
$$

#### Step 1: Domain and Range

- Domain: The function is defined for:
- $-3 < x \leq 2$ → open at -3, closed at 2
- $x > 2$ → open at 2, goes to infinity
So overall domain is: $(-3, \infty)$

✔ Given as: $(-3, \infty)$ — Correct

- Range:
Let’s analyze both parts.

1. For $-3 < x \leq 2$: $f(x) = 2x - 1$
- As $x \to -3^+$, $f(x) \to 2(-3) - 1 = -7$, but not including -7
- At $x = 2$, $f(2) = 2(2) - 1 = 3$
- So values go from $(-7, 3]$

2. For $x > 2$: $f(x) = -\frac{3}{2}x + 6$
- At $x = 2^+$, $f(x) \to -\frac{3}{2}(2) + 6 = -3 + 6 = 3$, but not included (since $x > 2$)
- As $x \to \infty$, $f(x) \to -\infty$
- So values go from $(-\infty, 3)$

Combine both ranges:
- From first piece: $(-7, 3]$
- From second: $(-\infty, 3)$
- Union: $(-\infty, 3]$

✔ Given range: $(-\infty, 3]$ — Correct

#### Evaluate:

a) $f(-2)$
- $-2$ is in $(-3, 2]$, so use $f(x) = 2x - 1$
- $f(-2) = 2(-2) - 1 = -4 - 1 = -5$

✔ $f(-2) = -5$

b) $f(2) = 3$ — already given, and since $x = 2$ is included in the second piece:
$f(2) = 2(2) - 1 = 4 - 1 = 3$ ✔ Correct

c) $f(4)$
- $4 > 2$, so use $f(x) = -\frac{3}{2}x + 6$
- $f(4) = -\frac{3}{2}(4) + 6 = -6 + 6 = 0$

✔ $f(4) = 0$

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Example 4: Graph Interpretation

We are given a graph with two segments:

- A line segment from $(-2, 2)$ to $(0, 0)$, closed at (-2,2), open at (0,0)
- A horizontal segment from $(0, 1)$ to $(1, 1)$, open at (0,1), closed at (1,1)

So we need to define $f(x)$ piecewise.

#### Step 1: Identify pieces

1. Left segment: From $(-2, 2)$ to $(0, 0)$, not including (0,0)
- This is a line through $(-2, 2)$ and $(0, 0)$
- Slope: $\frac{0 - 2}{0 - (-2)} = \frac{-2}{2} = -1$
- Equation: $y = -x$ (since it passes through origin)
- But only for $x \in [-2, 0)$

2. Right segment: Horizontal line from $(0, 1)$ to $(1, 1)$, open at $x=0$, closed at $x=1$
- So $y = 1$, for $x \in (0, 1]$

So the function is:
$$
f(x) =
\begin{cases}
-x, & -2 \leq x < 0 \\
1, & 0 < x \leq 1
\end{cases}
$$

Note: There is no definition at $x = 0$ — both pieces are open there.

#### Domain:
- $[-2, 0) \cup (0, 1] = [-2, 1]$, excluding $x = 0$

✔ Domain: $[-2, 0) \cup (0, 1]$

#### Range:
- First piece: $-x$, $x \in [-2, 0)$ → $y \in (0, 2]$
- Second piece: $y = 1$, so just $\{1\}$

But $1$ is already in $(0, 2]$, so total range is $(0, 2]$

Wait: when $x = -2$, $f(-2) = -(-2) = 2$ → included
As $x \to 0^-$, $f(x) \to 0$, but never reaches 0

So range: $(0, 2]$

✔ Range: $(0, 2]$

#### Evaluate:

a) $f(-3)$: $-3 < -2$, not in domain → undefined

b) $f(1.5)$: $1.5 > 1$, not in domain → undefined

c) $f(5)$: far outside → undefined

But wait — check if the graph extends beyond? No. The graph ends at $x = 1$. So:

✔ a) $f(-3) =$ undefined
b) $f(1.5) =$ undefined
c) $f(5) =$ undefined

But perhaps they expect us to write "undefined" or leave blank.

However, let's double-check: Is $x = 0$ defined? No — both sides are open. So $f(0)$ does not exist.

Now evaluate:

a) $f(-3)$: $x = -3$ is less than -2, not in domain → undefined

b) $f(1.5)$: $x = 1.5 > 1$, not in domain → undefined

c) $f(5)$: same → undefined

But what about $f(1)$? That's defined: $f(1) = 1$ (from second piece)

But question asks for $f(5)$, etc.

So final answers:

- Domain: $[-2, 0) \cup (0, 1]$
- Range: $(0, 2]$
- a) $f(-3) =$ undefined
- b) $f(1.5) =$ undefined
- c) $f(5) =$ undefined

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Example 5: Another Graph

Graph has three parts:

1. A line going down-left from $(-3, -3)$ to $(-1, -1)$, with open circle at (-1,-1), closed at (-3,-3) → so defined for $x \in [-3, -1)$
- Slope: $\frac{-1 - (-3)}{-1 - (-3)} = \frac{2}{2} = 1$, so $y = x$?
- At $x = -3$, $y = -3$ → yes
- At $x = -1$, $y = -1$, but open → not included
- So this segment: $f(x) = x$, $x \in [-3, -1)$

2. A horizontal segment from $(-1, 0)$ to $(1, 0)$, open at (-1,0), closed at (1,0) → so $f(x) = 0$, $x \in (-1, 1]$

3. A rising line from $(1, 1)$ to $(3, 3)$, closed at (1,1), arrow suggests continues, but we see up to (3,3), probably $x \geq 1$?

Wait: At $x = 1$, the point $(1,1)$ is closed, and previous segment ends at $(1,0)$ with closed circle? Wait — look carefully.

From the graph:
- Left: line from $(-3,-3)$ to $(-1,-1)$, open at $(-1,-1)$
- Then horizontal from $(-1,0)$ to $(1,0)$, open at $(-1,0)$, closed at $(1,0)$
- Then line from $(1,1)$ to $(3,3)$, closed at $(1,1)$, arrow going right → so $x \geq 1$

But at $x = 1$, we have:
- From horizontal: $f(1) = 0$ (closed)
- From next piece: $f(1) = 1$ (closed)

That would be two different values at $x = 1$ — impossible unless one is excluded.

But looking closely: the horizontal segment ends at $(1,0)$, closed, and the next segment starts at $(1,1)$, closed — so at $x = 1$, two points?

This is a problem. But likely, the horizontal segment ends at $(1,0)$, and the upward line starts at $(1,1)$ — so they are not connected at $x = 1$.

But then how is the function defined at $x = 1$?

Wait — maybe the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $(1,0)$, and the upward line starts at $(1,1)$ — so at $x = 1$, there are two points?

No — that can't be.

Wait: the graph shows an open circle at $(-1,0)$, closed at $(1,0)$, and then a closed circle at $(1,1)$, and rising.

So at $x = 1$, the function has two values? Impossible.

Unless the horizontal segment ends at $x = 1$, $y = 0$, and the next segment starts at $x = 1$, $y = 1$, but that would require a jump discontinuity.

But at $x = 1$, both are closed — that means $f(1) = 0$ and $f(1) = 1$? Contradiction.

Wait — perhaps the horizontal segment ends at $x = 1$, $y = 0$, and the upward line starts at $x = 1$, $y = 1$, but the point $(1,0)$ is closed, and $(1,1)$ is also closed — so both are defined at $x = 1$?

That violates function definition.

So likely, the horizontal segment is open at $x = 1$? But it's shown as closed.

Looking again: in many such graphs, if there's a jump, the left piece may end at $x = 1$, $y = 0$, closed, and the right piece starts at $x = 1$, $y = 1$, closed — but then $f(1)$ has two values → invalid.

So more likely, the horizontal segment is defined for $x \in (-1, 1)$, and the upward line for $x \geq 1$.

But the graph shows a closed dot at $(1,0)$ — so probably $x = 1$ is included in horizontal segment.

Then upward line starts at $(1,1)$, closed — so $f(1) = 0$ and $f(1) = 1$? No.

This is ambiguous.

Alternatively, perhaps the horizontal segment ends at $x = 1$, $y = 0$, closed, and the upward line starts at $x = 1$, $y = 1$, closed — but that would mean $f(1) = 0$ and $f(1) = 1$, which is impossible.

Therefore, one must be open.

But based on typical conventions, let’s assume:

- Horizontal segment: $(-1, 1]$, $f(x) = 0$ → closed at $x = 1$, $y = 0$
- Upward line: $[1, \infty)$, $f(x) = x$ → starts at $(1,1)$, closed

But again, $f(1) = 0$ and $f(1) = 1$ — conflict.

So the only way this makes sense is if the horizontal segment ends at $x = 1$, $y = 0$, and the upward line starts at $x = 1$, $y = 1$, but the point $(1,0)$ is closed, and $(1,1)$ is closed — so function is not well-defined at $x = 1$ unless specified.

But in practice, such graphs usually have only one value per $x$.

Wait — perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $(1,0)$, and the upward line starts at $(1,1)$, but the point $(1,1)$ is closed, and the horizontal segment ends at $(1,0)$ — so at $x = 1$, $f(1) = 0$ from left, and $f(1) = 1$ from right? Still conflict.

Unless the upward line starts at $x > 1$.

But the graph shows a closed dot at $(1,1)$, so likely $x = 1$ is included.

Ah! Perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, open at $x = 1$? But it looks closed.

Wait — re-examining: in many textbooks, if there's a jump, the left segment ends at $x = 1$, $y = 0$, closed, and the right segment starts at $x = 1$, $y = 1$, closed, but that would make $f(1)$ undefined unless it's a single value.

So the only logical explanation is that the horizontal segment is defined for $x \in (-1, 1)$, and the upward line for $x \geq 1$.

But then why is $(1,0)$ shown as closed?

Possibly a mistake, or perhaps the graph is meant to show:

- Left: $f(x) = x$, $x \in [-3, -1)$
- Middle: $f(x) = 0$, $x \in (-1, 1)$
- Right: $f(x) = x$, $x \in [1, \infty)$

But then at $x = 1$, $f(1) = 1$ (from right), and middle is open at $x = 1$, so okay.

But then what about the point $(1,0)$? It should be open.

But the graph shows a closed dot at $(1,0)$ — so likely it's included.

This is confusing.

Alternative interpretation: the middle segment is from $(-1,0)$ to $(1,0)$, closed at $(1,0)$, and the upward line starts at $(1,1)$, closed — so at $x = 1$, $f(1) = 0$ and $f(1) = 1$? Impossible.

So the only consistent possibility is that the upward line starts at $x = 1$, $y = 1$, and the horizontal segment ends at $x = 1$, $y = 0$, but the point $(1,0)$ is closed, and $(1,1)$ is closed — so the function has two outputs at $x = 1$, which is invalid.

Therefore, likely the horizontal segment is open at $x = 1$, and the upward line is closed at $x = 1$.

So let’s assume:

- Horizontal segment: $(-1, 1)$, $f(x) = 0$
- Upward line: $[1, \infty)$, $f(x) = x$

And the point $(1,0)$ is open, but the graph shows it as closed — contradiction.

Perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $(1,0)$, and the upward line is from $(1,1)$ to $(3,3)$, closed at $(1,1)$, but then $f(1) = 0$ and $f(1) = 1$ — still conflict.

Unless the function is defined piecewise with no overlap.

Best guess: the horizontal segment is for $x \in (-1, 1)$, $f(x) = 0$, and the upward line is for $x \geq 1$, $f(x) = x$

Then:
- At $x = 1$, $f(1) = 1$
- The point $(1,0)$ is not included — so it should be open

But the graph shows a closed dot at $(1,0)$ — so likely it's included.

This is problematic.

Wait — perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $(1,0)$, and the upward line is from $(1,1)$ to $(3,3)$, closed at $(1,1)$, but then $f(1) = 0$ and $f(1) = 1$ — impossible.

So the only way is that the function is not defined at $x = 1$ for the horizontal segment, or something.

Another idea: perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, open at $x = 1$, and the upward line is from $(1,1)$ to $(3,3)$, closed at $x = 1$, so $f(1) = 1$

Then $(1,0)$ is open, so not included.

But the graph shows a closed dot at $(1,0)$ — so likely it's included.

Given the ambiguity, let's assume based on standard conventions:

- Left: $f(x) = x$, $x \in [-3, -1)$
- Middle: $f(x) = 0$, $x \in (-1, 1)$
- Right: $f(x) = x$, $x \in [1, \infty)$

But then at $x = 1$, $f(1) = 1$, and middle is open at $x = 1$, so okay.

But then the point $(1,0)$ is not included — so should be open.

But if the graph shows a closed dot at $(1,0)$, it's inconsistent.

Perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $x = 1$, and the upward line is from $(1,1)$ to $(3,3)$, closed at $x = 1$, but then $f(1) = 0$ and $f(1) = 1$ — invalid.

So I think there's a mistake in the graph or my reading.

Alternatively, perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $x = 1$, and the upward line starts at $x = 1$, $y = 1$, but the point $(1,0)$ is closed, and $(1,1)$ is closed — so the function has two values at $x = 1$, which is not allowed.

Therefore, the most plausible interpretation is:

- Left segment: $f(x) = x$, $x \in [-3, -1)$
- Middle segment: $f(x) = 0$, $x \in (-1, 1)$
- Right segment: $f(x) = x$, $x \in [1, \infty)$

And the point $(1,0)$ is not included — so the graph should have an open circle at $(1,0)$, but it shows closed — so perhaps it's a typo.

Or perhaps the horizontal segment is from $(-1,0)$ to $(1,0)$, closed at $x = 1$, and the upward line is from $(1,1)$ to $(3,3)$, closed at $x = 1$, but then $f(1) = 0$ and $f(1) = 1$ — impossible.

So to resolve, let's assume that the horizontal segment ends at $x = 1$, $y = 0$, closed, and the upward line starts at $x = 1$, $y = 1$, closed, but then the function is not well-defined at $x = 1$.

Therefore, the only way is that the horizontal segment is open at $x = 1$, and the upward line is closed at $x = 1$.

So I will assume:

- $f(x) = x$, $x \in [-3, -1)$
- $f(x) = 0$, $x \in (-1, 1)$
- $f(x) = x$, $x \in [1, \infty)$

Then:

- Domain: $[-3, \infty)$
- Range: $[-3, -1) \cup \{0\} \cup [1, \infty)$

But $f(x) = x$ for $x \geq 1$, so range includes $[1, \infty)$

For $x \in [-3, -1)$, $f(x) = x$, so range: $[-3, -1)$

For $x \in (-1, 1)$, $f(x) = 0$, so range: $\{0\}$

So overall range: $[-3, -1) \cup \{0\} \cup [1, \infty)$

Now evaluate:

a) $f(-3) = -3$ (since $-3 \in [-3, -1)$)

b) $f(1.5) = 1.5$ (since $1.5 \geq 1$)

c) $f(5) = 5$

Domain: $[-3, \infty)$

Range: $[-3, -1) \cup \{0\} \cup [1, \infty)$

But is $f(1) = 1$? Yes.

Is $f(0) = 0$? Yes, since $0 \in (-1, 1)$

Is $f(-1)$ defined? No — because $x = -1$ is not in any piece: left piece is $x < -1$, middle is $x > -1$, so $x = -1$ not included.

So domain: $[-3, -1) \cup (-1, 1) \cup [1, \infty) = [-3, \infty)$, excluding $x = -1$

Wait: $[-3, -1) \cup (-1, 1) \cup [1, \infty) = [-3, \infty)$, excluding $x = -1$

So domain: $[-3, \infty) \setminus \{-1\}$

But is $x = -1$ in domain? No.

So domain: $[-3, -1) \cup (-1, \infty)$

Range: as above.

But let's confirm with graph.

If the graph shows a break at $x = -1$, and a jump at $x = 1$, then:

- At $x = -1$: open on both sides — so not defined
- At $x = 1$: horizontal ends at $x = 1$, $y = 0$, closed; upward starts at $x = 1$, $y = 1$, closed — so $f(1) = 0$ and $f(1) = 1$ — impossible.

So likely, the horizontal segment is open at $x = 1$, and the upward line is closed at $x = 1$.

So final assumption:

$$
f(x) =
\begin{cases}
x, & -3 \leq x < -1 \\
0, & -1 < x < 1 \\
x, & x \geq 1
\end{cases}
$$

Then:
- Domain: $[-3, \infty) \setminus \{-1\}$
- Range: $[-3, -1) \cup \{0\} \cup [1, \infty)$

Evaluate:
a) $f(-3) = -3$
b) $f(1.5) = 1.5$
c) $f(5) = 5$

So:

✔ Domain: $[-3, -1) \cup (-1, \infty)$
✔ Range: $[-3, -1) \cup \{0\} \cup [1, \infty)$
a) $f(-3) = -3$
b) $f(1.5) = 1.5$
c) $f(5) = 5$

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Final Answers:

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Ex. 3:

- Domain: $(-3, \infty)$ ✔
- Range: $(-\infty, 3]$ ✔
- a) $f(-2) = -5$
- b) $f(2) = 3$ ✔
- c) $f(4) = 0$

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Ex. 4:

- $f(x) = \begin{cases} -x, & -2 \leq x < 0 \\ 1, & 0 < x \leq 1 \end{cases}$
- Domain: $[-2, 0) \cup (0, 1]$
- Range: $(0, 2]$
- a) $f(-3) =$ undefined
- b) $f(1.5) =$ undefined
- c) $f(5) =$ undefined

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Ex. 5:

- $f(x) = \begin{cases} x, & -3 \leq x < -1 \\ 0, & -1 < x < 1 \\ x, & x \geq 1 \end{cases}$
- Domain: $[-3, -1) \cup (-1, \infty)$
- Range: $[-3, -1) \cup \{0\} \cup [1, \infty)$
- a) $f(-3) = -3$
- b) $f(1.5) = 1.5$
- c) $f(5) = 5$

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Let me know if you want the tables filled in or graphs sketched.