Math Problem Statement

The solid points denote the value of the function at the points, and the values denoted by the hollow points are not taken by the functions.

The solid points denote the value of the function at the points, and the values denoted by the hollow points are not taken by the functions.

Solution

Let us analyze the problem and justify the selected options based on the given graph:

  1. Limits at t2t \to 2^- and t2+t \to 2^+:

    • As tt approaches 2 from the left (t2t \to 2^-), the function f(t)f(t) approaches the value 10. This matches the graph where the left-hand limit is 10 at t=2t = 2.
    • As tt approaches 2 from the right (t2+t \to 2^+), the function f(t)f(t) approaches 2. This also aligns with the graph, showing the right-hand limit is 2 at t=2t = 2.

    Hence, limt2f(t)=10\lim_{t \to 2^-} f(t) = 10 and limt2+f(t)=2\lim_{t \to 2^+} f(t) = 2 are correct.

  2. Continuity at t=2t = 2:

    • A function is continuous at t=2t = 2 if limt2f(t)=f(2)\lim_{t \to 2} f(t) = f(2). Here, the left-hand limit and right-hand limit are not equal (10210 \neq 2), so the two-sided limit limt2f(t)\lim_{t \to 2} f(t) does not exist. Therefore, the function is not continuous at t=2t = 2.

    Thus, ff is not continuous at t=2t = 2.

  3. Limits at t4t \to 4^- and t4+t \to 4^+:

    • As tt approaches 4 from the left (t4t \to 4^-), the function f(t)f(t) approaches a value of 2. However, the limit at t4t \to 4^- is marked as false because it does not match the expected value of 4.
    • As tt approaches 4 from the right (t4+t \to 4^+), the function f(t)f(t) approaches 4. This is correct as per the graph.

    Therefore, limt4+f(t)=4\lim_{t \to 4^+} f(t) = 4 is correct.

  4. Continuity at t=6t = 6:

    • At t=6t = 6, the graph shows a straight line segment with no breaks or jumps, indicating that the function is continuous at t=6t = 6.

    Thus, ff is continuous at t=6t = 6.

Summary of Correct Options:

  • limt2f(t)=10\lim_{t \to 2^-} f(t) = 10 (Correct)
  • limt2+f(t)=2\lim_{t \to 2^+} f(t) = 2 (Correct)
  • limt4f(t)=4\lim_{t \to 4^-} f(t) = 4 (Incorrect)
  • limt4+f(t)=4\lim_{t \to 4^+} f(t) = 4 (Correct)
  • ff is continuous at t=2t = 2 (Incorrect)
  • ff is continuous at t=6t = 6 (Correct)

Let me know if you'd like further clarification or additional examples!


Here are 5 related questions to explore:

  1. How do we calculate one-sided and two-sided limits algebraically?
  2. Can a function have a left-hand limit and right-hand limit but still be discontinuous?
  3. What types of discontinuities exist in functions (e.g., jump, removable)?
  4. How can a piecewise function be defined to ensure continuity at all points?
  5. Why is continuity important in calculus and real-world applications?

Tip: Always check both the left-hand and right-hand limits at a point to determine continuity!

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Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Piecewise Functions

Formulas

Left-hand limit: \( \lim_{x \to c^-} f(x) \)
Right-hand limit: \( \lim_{x \to c^+} f(x) \)
Continuity condition: \( \lim_{x \to c} f(x) = f(c) \)

Theorems

Definition of continuity
Properties of limits

Suitable Grade Level

Grades 11-12 or College Level