Math Problem Statement

Знайти та класифікувати точки розриву функції \(y = (1 + x) \arctan \frac{1}{1 - x^2}\). Намалювати поведінку графіку функції поблизу точок розриву.

Solution

The task is to identify and classify the points of discontinuity of the given function:

y=(1+x)arctan(11x2)y = (1 + x) \arctan\left(\frac{1}{1 - x^2}\right)

and analyze the behavior of the function's graph near those points.

Step 1: Analyze the function for discontinuities

  1. The term 11x2\frac{1}{1 - x^2} introduces potential discontinuities because it becomes undefined when 1x2=01 - x^2 = 0, i.e., x=±1x = \pm 1.
  2. The function arctan(z)\arctan(z) is defined for all real zz, so the only concern is the discontinuity of 11x2\frac{1}{1 - x^2}.

Thus, the points of potential discontinuity are x=±1x = \pm 1.


Step 2: Classify the discontinuities at x=±1x = \pm 1

  • Asymptotic behavior: Near x=±1x = \pm 1, the denominator 1x21 - x^2 approaches 00, leading to 11x2±\frac{1}{1 - x^2} \to \pm \infty.
    • As x1x \to 1^-, 1x20+1 - x^2 \to 0^+, so 11x2+\frac{1}{1 - x^2} \to +\infty.
    • As x1+x \to 1^+, 1x201 - x^2 \to 0^-, so 11x2\frac{1}{1 - x^2} \to -\infty.
    • Similarly, for x1x \to -1, the behavior is symmetric.

Thus, the points x=±1x = \pm 1 are vertical asymptotes of the function, indicating infinite discontinuities.


Step 3: Behavior near the discontinuities

  1. Near x=1x = 1:
    • As x1x \to 1^-, (1+x)arctan(11x2)(1+1)arctan(+)=2π2=π(1 + x) \arctan\left(\frac{1}{1 - x^2}\right) \to (1 + 1) \cdot \arctan(+\infty) = 2 \cdot \frac{\pi}{2} = \pi.
    • As x1+x \to 1^+, (1+x)arctan(11x2)(1+1)arctan()=2(π2)=π(1 + x) \arctan\left(\frac{1}{1 - x^2}\right) \to (1 + 1) \cdot \arctan(-\infty) = 2 \cdot \left(-\frac{\pi}{2}\right) = -\pi.

Thus, there is a jump discontinuity at x=1x = 1.

  1. Near x=1x = -1:
    • As x1x \to -1^-, (1+x)arctan(11x2)0arctan(+)=0(1 + x) \arctan\left(\frac{1}{1 - x^2}\right) \to 0 \cdot \arctan(+\infty) = 0.
    • As x1+x \to -1^+, (1+x)arctan(11x2)0arctan()=0(1 + x) \arctan\left(\frac{1}{1 - x^2}\right) \to 0 \cdot \arctan(-\infty) = 0.

Thus, x=1x = -1 is a removable discontinuity because the function's value approaches 00 from both sides.


Step 4: Graph behavior near discontinuities

  1. At x=1x = 1:

    • The function has a jump discontinuity, with the left limit approaching π\pi and the right limit approaching π-\pi.
  2. At x=1x = -1:

    • The function smoothly approaches 00 from both sides, indicating a removable discontinuity.

Would you like a detailed plot of the graph to visualize these behaviors? Let me know if you need further clarification or numerical calculations.


Follow-up Questions:

  1. Would you like a step-by-step explanation of the arctan\arctan function's role in the discontinuities?
  2. Should I compute additional limits for other points of interest?
  3. Do you need clarification on removable vs. infinite discontinuities?
  4. Would you like to analyze other features of the graph, such as symmetry or periodicity?
  5. Should I include an extended explanation on asymptotic behavior near vertical asymptotes?

Tip:

When analyzing discontinuities, always check the behavior of both the numerator and denominator to fully understand the function's behavior near the critical points.

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Math Problem Analysis

Mathematical Concepts

Analysis
Discontinuities
Asymptotes
Limits

Formulas

\(y = (1 + x) \arctan \frac{1}{1 - x^2}\)

Theorems

Limit behavior at points of discontinuity
Classification of discontinuities

Suitable Grade Level

University level