Math Problem Statement

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.

Solution

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)(-3, 0), increases as it moves to the right, and ends at the open point (1,2)(-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)(-1, 1), suggesting that the function is defined at x=1x = -1 with value 1.
  • The curve then moves downward and right, crossing the x-axis at x=0.5x = -0.5, and the y-axis at y=0.75y = -0.75.
  • The function changes direction at (0.5,1.9)(0.5, -1.9), then moves upward, crossing the x-axis at x=1.2x = 1.2, and exits in the first quadrant.

Continuity at x=1x = -1:

The key issue is at x=1x = -1, where the function transitions between the two parts:

  • Left-hand limit (as x1x \to -1^-): From the description of the first part, the function approaches (1,2)(-1, 2) from the left, which means: limx1f(x)=2\lim_{x \to -1^-} f(x) = 2

  • Right-hand limit (as x1+x \to -1^+): From the description of the second part, the function starts at the closed point (1,1)(-1, 1), so: limx1+f(x)=1\lim_{x \to -1^+} f(x) = 1

Since the left-hand limit limx1f(x)=2\lim_{x \to -1^-} f(x) = 2 is not equal to the right-hand limit limx1+f(x)=1\lim_{x \to -1^+} f(x) = 1, the limit at x=1x = -1 does not exist. Therefore, the function is not continuous at x=1x = -1.

Conclusion:

The function is not continuous at x=1x = -1 because: limx1f(x)limx1+f(x)\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Continuity
Limits
Piecewise Functions
Graph Interpretation

Formulas

lim x→c f(x)
Continuity condition: lim x→c⁻ f(x) = lim x→c⁺ f(x) = f(c)

Theorems

Theorem of Continuity at a Point
Limit of a Function

Suitable Grade Level

Grades 10-12