https://file.aiquickdraw.com/mathbot/file/17276331785462so7dz2f.jpg

The image contains multiple mathematical functions, and the task is to determine at which points each function is continuous. I will analyze each function and explain where they may fail to be continuous.

1. **i) \( y = \frac{1}{x - 2} - 3x \)**
   - This is a rational function. It is continuous everywhere except where the denominator is zero.
   - **Discontinuity**: \( x = 2 \).

2. **ii) \( y = \frac{1}{(x + 2)^2} + 4 \)**
   - This is also a rational function. It is continuous except where the denominator is zero.
   - **Discontinuity**: \( x = -2 \) (since \( (x + 2)^2 = 0 \) at \( x = -2 \)).

3. **iii) \( y = \frac{x + 1}{x^2 - 4x + 3} \)**
   - The denominator factors as \( (x - 3)(x - 1) \), so the function is continuous except where the denominator is zero.
   - **Discontinuities**: \( x = 1 \) and \( x = 3 \).

4. **iv) \( y = \frac{x + 3}{x^2 - 3x + 10} \)**
   - The denominator \( x^2 - 3x + 10 \) has no real roots because the discriminant \( (-3)^2 - 4(1)(10) = -31 \) is negative.
   - **No discontinuities**, continuous for all real \( x \).

5. **v) \( y = |x - 1| + \sin(x) \)**
   - The absolute value function and sine function are continuous everywhere.
   - **Continuous everywhere**.

6. **vi) \( y = \frac{1}{|x| + 1} - \frac{x^2}{2} \)**
   - The first term \( \frac{1}{|x| + 1} \) is continuous because \( |x| + 1 \neq 0 \) for all real \( x \). The second term \( \frac{x^2}{2} \) is a polynomial, so it's also continuous everywhere.
   - **Continuous everywhere**.

7. **vii) \( y = \frac{\cos(x)}{x} \)**
   - This is a rational function. It is continuous everywhere except where the denominator is zero.
   - **Discontinuity**: \( x = 0 \).

8. **viii) \( y = \frac{x + 2}{\cos(x)} \)**
   - This is also a rational function. It is continuous except where the denominator \( \cos(x) = 0 \), which happens at \( x = \frac{\pi}{2} + n\pi \) for integers \( n \).
   - **Discontinuities**: \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \).

9. **ix) \( g(x) = \begin{cases} x^2 - x - 6 & x \neq 3 \\ 5 & x = 3 \end{cases} \)**
   - The function is piecewise. For \( x \neq 3 \), the function is \( x^2 - x - 6 = (x - 3)(x + 2) \), which is continuous for all \( x \). However, at \( x = 3 \), the function takes the value 5, which is different from the value of \( (x - 3)(x + 2) \) at \( x = 3 \), which would be \( 0 \).
   - **Discontinuity**: \( x = 3 \).

### Summary of Discontinuities:
- i) \( x = 2 \)
- ii) \( x = -2 \)
- iii) \( x = 1, x = 3 \)
- iv) None
- v) None
- vi) None
- vii) \( x = 0 \)
- viii) \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \)
- ix) \( x = 3 \)

Would you like more details on any of these functions or their continuity? Here are five related questions to further explore:

1. What does the Intermediate Value Theorem say about continuous functions?
2. How do we determine the type of discontinuity (removable, jump, etc.)?
3. How does continuity affect the differentiability of a function?
4. Why are rational functions often discontinuous at specific points?
5. Can we make a discontinuous function continuous by redefining a point?

**Tip**: The absolute value function \( |x| \) is continuous everywhere, but its derivative has a sharp corner at \( x = 0 \), making it non-differentiable at that point.

At what points are the following functions continuous?

This problem explores how to determine points of continuity in various types of functions, including rational, trigonometric, and piecewise functions. Learn how to identify where these functions may have discontinuities and what points remain continuous.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation