The graph of a function f is given.

The x y-coordinate plane is given. A curve with 2 parts and 2 points is graphed.
The first part enters the window in the third quadrant, goes up and right, changes direction at the open point (−1, −1) below the first point (−1, 1), goes down and right, and exits the window just to the left of the y-axis.
The second point is at (0, 2).
The second part enters the window just to the right of the positive y-axis, goes down and right, crosses the x-axis at x = 1, and exits the window in the fourth quadrant.
Determine whether f is continuous on its domain.
continuous
not continuous    
Incorrect: Your answer is incorrect.
If it is not continuous on its domain, say why.

lim 
x→0+ 
f(x) ≠
lim 
x→0− 
f(x), so
lim 
x→0 
f(x) does not exist.
The graph has discontinuities at the end points.    

lim 
x→−1 
f(x) = −1 ≠ f(−1) and 
lim 
x→0 
f(x) does not exist.
The graph is continuous on its domain.
Incorrect: Your answer is incorrect.


To determine whether the function \( f(x) \) is continuous on its domain, we need to examine both the graphical behavior and the formal definition of continuity. A function is continuous on its domain if, at every point within the domain, the limit of the function as \( x \) approaches the point from both directions exists and is equal to the function's value at that point.

### Let's break down the graph:

1. **Point at \( x = -1 \)**:
   - The graph shows an open circle at \( (-1, -1) \) and a closed point at \( (-1, 1) \).
   - The left-hand limit as \( x \to -1^- \) (from the left) is \( \lim_{x \to -1^-} f(x) = -1 \), as the graph approaches \( (-1, -1) \).
   - The right-hand limit as \( x \to -1^+ \) (from the right) is \( \lim_{x \to -1^+} f(x) = -1 \).
   - However, \( f(-1) = 1 \), based on the closed point at \( (-1, 1) \). Since \( f(-1) \neq \lim_{x \to -1} f(x) \), the function is **discontinuous at \( x = -1 \)**.

2. **Point at \( x = 0 \)**:
   - The function has no value at \( x = 0 \) (there is no defined point there), but there are two different behaviors on either side:
     - As \( x \to 0^+ \) (from the right), the graph approaches a value of \( 2 \).
     - As \( x \to 0^- \) (from the left), the graph approaches a value near \( -1 \).
   - Since \( \lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x) \), the limit does not exist at \( x = 0 \). This makes the function **discontinuous at \( x = 0 \)**.

### Conclusion:

The function \( f(x) \) is **not continuous on its domain**. It is discontinuous at \( x = -1 \) because \( \lim_{x \to -1} f(x) \neq f(-1) \), and at \( x = 0 \) because \( \lim_{x \to 0} f(x) \) does not exist.

Would you like more details on any specific aspect? Here are some related questions:

1. What is the formal definition of continuity at a point?
2. How can you find limits graphically?
3. What are the types of discontinuities?
4. How would the function look if it were continuous?
5. What is the difference between removable and non-removable discontinuities?

**Tip**: When checking for continuity, always compare the left-hand limit, right-hand limit, and the function's value at the point!

The graph of a function f is given. The x y-coordinate plane is given. A curve with 2 parts and 2 points is graphed. The first part enters the window in the third quadrant, goes up and right, changes direction at the open point (−1, −1) below the first point (−1, 1), goes down and right, and exits the window just to the left of the y-axis. The second point is at (0, 2). The second part enters the window just to the right of the positive y-axis, goes down and right, crosses the x-axis at x = 1, and exits the window in the fourth quadrant. Determine whether f is continuous on its domain. continuous not continuous. Incorrect: Your answer is incorrect. If it is not continuous on its domain, say why. lim x→0+ f(x) ≠ lim x→0− f(x), so lim x→0 f(x) does not exist. The graph has discontinuities at the end points. lim x→−1 f(x) = −1 ≠ f(−1) and lim x→0 f(x) does not exist. The graph is continuous on its domain. Incorrect: Your answer is incorrect.

Analyze the continuity of a function based on its graph, with a focus on discontinuities at x = -1 and x = 0. Learn key concepts like limits, continuity conditions, and piecewise functions, including why the limit does not exist at certain points. Suitable for Grades 11-12.

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