Math Problem Statement

Which of the following is a function that has a jump discontinuity at x = 2 and a removable discontinuity at x = 4, but is continuous elsewhere?

Solution

The question asks to identify which of the three given functions has a jump discontinuity at x=2x = 2 and a removable discontinuity at x=4x = 4, while being continuous elsewhere.

Let's examine the functions one by one:

Option (a):

f(x)=2(x−2)(x−4)f(x) = \frac{2}{(x-2)(x-4)}

  • This function has potential discontinuities at x=2x = 2 and x=4x = 4** because those values make the denominator zero.
  • The discontinuity at x=2x = 2 is a non-removable discontinuity (likely a vertical asymptote, since the numerator remains 2).
  • The discontinuity at x=4x = 4 is removable because the denominator can be factored to (x−4)(x-4), and thus can be simplified.

This matches the condition of having a removable discontinuity at x=4x = 4 and a jump discontinuity at x=2x = 2.

Option (b):

1 & \text{if } x \leq 2 \\ x - 3 & \text{if } 2 < x < 4 \text{ or } x > 4 \\ 3 & \text{if } x = 4 \end{cases}$$ - At $$x = 2$$, the function is continuous because it is defined as $$f(x) = 1$$ for $$x \leq 2$$, and it smoothly connects for values larger than 2 (since $$x-3$$ is continuous in its interval). - At $$x = 4$$, the function jumps from $$f(x) = 1$$ to $$f(x) = 3$$, indicating a **jump discontinuity**. This doesn't match the condition because there's no removable discontinuity at $$x = 4$$. ### Option (c): $$f(x) = \begin{cases} 2 - x^2 & \text{if } x \leq 2 \\ \frac{1}{x^2 - 4x} & \text{if } x > 2 \end{cases}$$ - The function has a **potential discontinuity** at $$x = 2$$, but since the left-hand limit and right-hand limit do not match, it indicates a **jump discontinuity** at $$x = 2$$. - The denominator $$x^2 - 4x = x(x - 4)$$ suggests a discontinuity at $$x = 4$$. The discontinuity at $$x = 4$$ is **non-removable** since it's in the denominator. This option does not satisfy the condition either. ### Conclusion: The correct answer is **(a)** because it has a **jump discontinuity at $$x = 2$$** and a **removable discontinuity at $$x = 4$$**, while being continuous elsewhere. Would you like further details on this solution? Here are some related questions to explore: 1. What is the difference between removable and non-removable discontinuities? 2. How can we classify discontinuities based on the behavior of limits? 3. Why is a jump discontinuity non-removable? 4. How do you find removable discontinuities by factoring? 5. Can a function be continuous if it has a removable discontinuity? **Tip:** Always check both the numerator and the denominator when identifying discontinuities in rational functions!

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Math Problem Analysis

Mathematical Concepts

Discontinuities
Jump Discontinuity
Removable Discontinuity
Rational Functions

Formulas

f(x) = 2 / ((x-2)(x-4))

Theorems

Discontinuity Classification

Suitable Grade Level

Grades 11-12 (Pre-Calculus or Calculus)

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