3.3: Domain and Range - Mathematics LibreTexts - Free Printable

It looks like you've uploaded a graph of a function $ f(x) $, and I need to analyze it. Let's go through the details of the graph and solve any problem that might be associated with it.

Observations from the Graph:

- The graph is defined on the interval $ (-3, 1] $.
- There is an open circle at $ x = -3 $, meaning $ x = -3 $ is not included in the domain.
- There is a closed circle at $ x = 1 $, meaning $ x = 1 $ is included.
- The function appears to be continuous on $ (-3, 1] $.
- The function has:
- A local minimum near $ x = -2 $, where $ y \approx -4 $.
- A local maximum at $ x = 0 $, where $ y = 0 $.
- The function starts at $ x \to -3^+ $, with $ y \to -1 $ (since the open circle is at $ (-3, -1) $).
- At $ x = 1 $, $ y = -4 $, so $ f(1) = -4 $.

---

Now, since no specific question was stated, I’ll assume a common type of problem associated with such graphs: Find the range of the function $ f $.

---

Step-by-step Solution: Find the Range of $ f $

The range is the set of all possible output values ($ y $-values) that the function takes on over its domain.

From the graph:

- The lowest value of $ f(x) $ is $ -4 $, which occurs at both $ x = -2 $ and $ x = 1 $.
- The highest value is $ 0 $, which occurs at $ x = 0 $.
- The function is continuous between $ x = -3 $ and $ x = 1 $, and covers all values between $ -4 $ and $ 0 $.
- At $ x \to -3^+ $, $ f(x) \to -1 $, but does not include $ f(-3) $, so $ -1 $ is approached but not necessarily achieved — however, looking closely, the function seems to reach $ y = -1 $ just after $ x = -3 $, but since the point is open at $ (-3, -1) $, we must check if $ y = -1 $ is attained elsewhere.

Wait — actually, let’s examine this more carefully.

At $ x \to -3^+ $, the function approaches $ y = -1 $, but since it's an open circle, $ f(x) $ never actually reaches $ y = -1 $ at $ x = -3 $. However, as $ x $ increases slightly beyond $ -3 $, the function dips down to $ y = -4 $ at $ x = -2 $, then rises to $ y = 0 $ at $ x = 0 $, and finally drops to $ y = -4 $ again at $ x = 1 $.

So, what are the actual $ y $-values?

- The function goes from just above $ y = -1 $ at $ x \to -3^+ $, down to $ y = -4 $ at $ x = -2 $, up to $ y = 0 $ at $ x = 0 $, then down to $ y = -4 $ at $ x = 1 $.
- So, the maximum value is $ 0 $.
- The minimum value is $ -4 $.
- Since the function is continuous on $ (-3, 1] $, and it reaches $ y = -4 $ and $ y = 0 $, and everything in between is covered (by Intermediate Value Theorem), the range includes all values from $ -4 $ to $ 0 $.

But wait: at $ x \to -3^+ $, $ y \to -1 $, but does it ever go below $ -1 $? Yes — it goes down to $ -4 $. So the function achieves values from $ -4 $ up to $ 0 $.

But does it achieve $ y = -1 $? Yes — for example, somewhere between $ x = -3 $ and $ x = -2 $, the function passes through $ y = -1 $, even though it doesn’t include the point at $ x = -3 $. So $ y = -1 $ is in the range.

Also, since the function reaches $ y = 0 $ at $ x = 0 $, and $ y = -4 $ at $ x = -2 $ and $ x = 1 $, and is continuous, the entire interval from $ -4 $ to $ 0 $ is covered.

So the range is all real numbers from $ -4 $ to $ 0 $, inclusive.

---

✔ Final Answer:

$$
\boxed{[-4,\ 0]}
$$

This is the range of the function $ f $.

---

If you had a different question in mind (e.g., find $ f(-1) $, determine intervals of increase/decrease, or find limits), please let me know and I’ll adjust accordingly!