https://file.aiquickdraw.com/mathbot/file/1727659668129bjzi61v8.jpg

Let's analyze the given piecewise function:

\[
w(x) = \begin{cases} 
x + 1 & \text{if } x \leq 0 \\
\frac{1}{5}x^2 + 1 & \text{if } x > 0
\end{cases}
\]

We need to check for discontinuities at \(x = 0\), which is where the behavior of the function changes. Here's how we proceed:

### 1. **Check if the function is continuous at \(x = 0\)**:
   - The left-hand limit (as \(x \to 0^-\)) of the function comes from the first piece \(x + 1\):
     \[
     \lim_{x \to 0^-} w(x) = 0 + 1 = 1
     \]

   - The right-hand limit (as \(x \to 0^+\)) comes from the second piece \(\frac{1}{5}x^2 + 1\):
     \[
     \lim_{x \to 0^+} w(x) = \frac{1}{5}(0^2) + 1 = 1
     \]

   - The value of the function at \(x = 0\) (from the first piece, since \(x = 0 \leq 0\)) is:
     \[
     w(0) = 0 + 1 = 1
     \]

### 2. **Conclusion**:
   Since both the left-hand limit, right-hand limit, and the function's value at \(x = 0\) are equal, \(w(x)\) is continuous at \(x = 0\).

### Answer:
- There are **no discontinuities** for this function. Thus, you should write **None** for both the removable and non-removable discontinuities.

Would you like any further clarification on this problem?

---

Here are some additional related questions for practice:

1. What is the definition of a removable discontinuity?
2. How do you determine if a discontinuity is non-removable?
3. Can a piecewise function have discontinuities at points other than where the pieces join?
4. What are the conditions for a function to be continuous at a point?
5. How does the concept of limits help in identifying continuity in piecewise functions?

**Tip**: Always check the behavior of a piecewise function at the points where the pieces change by calculating the limits from both sides and the function's value at that point.

Find and classify the discontinuities of the following function as removable or nonremovable. If a classification has no discontinuities, write None for your answer. The function is: w(x) = { x + 1 if x <= 0; (1/5)x^2 + 1 if x > 0 }. Removable discontinuity: c = ; Nonremovable discontinuity: c = 

Analyze the given piecewise function to find and classify discontinuities at x = 0. Learn how to use limits from both sides and evaluate function continuity. This problem covers key concepts in limits and continuity, suitable for high school students or Calculus I learners.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation