https://file.aiquickdraw.com/mathbot/file/1723675429294c10dv184.jpg

To determine where the function \( k(x) \) is continuous, let's analyze the piecewise definition provided in the image:

\[
k(x) = \begin{cases} 
\frac{x-2}{x^3 - 8} & \text{if } x \neq 2 \\
12 & \text{if } x = 2 
\end{cases}
\]

### Steps to determine continuity:

1. **Continuity for \( x \neq 2 \):**  
   The function \( k(x) \) is continuous for all \( x \) except possibly at \( x = 2 \). The first piece \( \frac{x-2}{x^3 - 8} \) is continuous wherever it is defined.

   Notice that \( x^3 - 8 = (x - 2)(x^2 + 2x + 4) \). Hence, the function can be rewritten as:
   \[
   \frac{x-2}{(x-2)(x^2 + 2x + 4)} = \frac{1}{x^2 + 2x + 4}
   \]
   for \( x \neq 2 \).

   Since \( x^2 + 2x + 4 \) is always positive and never zero for all real \( x \), the function is continuous for all \( x \neq 2 \).

2. **Continuity at \( x = 2 \):**  
   For the function to be continuous at \( x = 2 \), the following must hold:
   \[
   \lim_{x \to 2} k(x) = k(2)
   \]
   We already know \( k(2) = 12 \).

   Now, let's compute the limit as \( x \) approaches 2:
   \[
   \lim_{x \to 2} \frac{1}{x^2 + 2x + 4} = \frac{1}{2^2 + 2(2) + 4} = \frac{1}{4 + 4 + 4} = \frac{1}{12}
   \]

   Since the limit as \( x \) approaches 2 is \( \frac{1}{12} \) and \( k(2) = 12 \), these are not equal. Therefore, \( k(x) \) is **not** continuous at \( x = 2 \).

### Conclusion:
The function \( k(x) \) is continuous for all \( x \) except at \( x = 2 \).

Would you like to go over the concepts in more detail or ask any questions?

Here are some related questions for further exploration:

1. What is the definition of continuity at a point?
2. How do you find the limit of a piecewise function at a point where the pieces meet?
3. Why does the factorization of \( x^3 - 8 \) simplify the function \( k(x) \)?
4. What happens to the continuity if \( k(x) \) were defined differently at \( x = 2 \)?
5. Can you give an example of a function that is continuous everywhere?
6. What is the difference between continuity and differentiability?
7. How can a function be made continuous by redefining a specific point?
8. What role does the intermediate value theorem play in analyzing the continuity of functions?

**Tip:** When checking for continuity, always examine the function at the points where the definition changes or where the function might not be well-defined.

Explore how to determine the continuity of a piecewise function using detailed analysis and examples. This problem involves understanding limits, continuity at specific points, and piecewise function behavior. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation