The graph of a function f is given.

The x y-coordinate plane is given. A curve with 2 parts is graphed.
The first part is linear, enters the window at the point (−3, 0), goes up and right, and ends at the open point (−1, 2).
The second part begins at the closed point (−1, 1), goes down and right, crosses the x-axis at x = −0.5, crosses the y-axis at y = −0.75, changes direction at (0.5, −1.9), goes up and right, crosses the x-axis at x = 1.2, and exits the window in the first quadrant.
Determine whether f is continuous on its domain.
continuous
not continuous    
If it is not continuous on its domain, say why.

lim 
x→−1+ 
f(x) ≠
lim 
x→−1− 
f(x), so
lim 
x→−1 
f(x) does not exist.
The function is not defined at x = −1.    

lim 
x→−1+ 
f(x) ≠ f(−1)
The graph is continuous on its domain.

From the description of the graph, let's break down the key observations:

### 1. First part of the graph:
- The curve is **linear** and enters at the point \((-3, 0)\), increases as it moves to the right, and ends at the **open point** \((-1, 2)\). 
  - Since it’s a straight line and the function is defined everywhere along this segment, we know that the function is continuous over this section of the domain.

### 2. Second part of the graph:
- This part starts from the **closed point** \((-1, 1)\), suggesting that the function is defined at \(x = -1\) with value 1.
- The curve then moves downward and right, crossing the x-axis at \(x = -0.5\), and the y-axis at \(y = -0.75\).
- The function changes direction at \((0.5, -1.9)\), then moves upward, crossing the x-axis at \(x = 1.2\), and exits in the first quadrant.

### Continuity at \(x = -1\):
The key issue is **at \(x = -1\)**, where the function transitions between the two parts:

- **Left-hand limit** (as \(x \to -1^-\)): From the description of the first part, the function approaches \((-1, 2)\) from the left, which means:
  \[
  \lim_{x \to -1^-} f(x) = 2
  \]
  
- **Right-hand limit** (as \(x \to -1^+\)): From the description of the second part, the function starts at the closed point \((-1, 1)\), so:
  \[
  \lim_{x \to -1^+} f(x) = 1
  \]

Since the **left-hand limit** \(\lim_{x \to -1^-} f(x) = 2\) is not equal to the **right-hand limit** \(\lim_{x \to -1^+} f(x) = 1\), the limit at \(x = -1\) does not exist. Therefore, the function is **not continuous** at \(x = -1\).

### Conclusion:
The function is **not continuous** at \(x = -1\) because:
\[
\lim_{x \to -1^-} f(x) \neq \lim_{x \to -1^+} f(x)
\]
Thus, the function is **not continuous on its domain**.

---

**Would you like further details or an explanation on how to handle limits or continuity at specific points?**

Here are some follow-up questions to dive deeper:

1. What is the definition of continuity for a function?
2. How do you evaluate limits from the left and right of a point?
3. What is the difference between open and closed points on a graph?
4. How do you determine if a function is continuous at a specific point or interval?
5. What happens to the continuity of a function if the limits at a point do not match?

**Tip:** To check for continuity at a point, always ensure that the left-hand limit, right-hand limit, and the function's value at the point all match.

The graph of a function f is given. The x y-coordinate plane is given. A curve with 2 parts is graphed. The first part is linear, enters the window at the point (−3, 0), goes up and right, and ends at the open point (−1, 2). The second part begins at the closed point (−1, 1), goes down and right, crosses the x-axis at x = −0.5, crosses the y-axis at y = −0.75, changes direction at (0.5, −1.9), goes up and right, crosses the x-axis at x = 1.2, and exits the window in the first quadrant. Determine whether f is continuous on its domain.

Learn how to determine the continuity of a piecewise function at a given point using the concept of limits. This problem involves evaluating the limits from both sides of x = -1 to determine continuity. Suitable for students in Grades 10-12 studying calculus or advanced algebra.

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