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The problem asks to determine if each function is continuous at the given \( x \)-values. If not continuous, the discontinuity should be classified.

Let's analyze each function one by one.

### 1a) \( f(x) = \frac{x^2 - 1}{x - 1} \) at \( x = 1 \)

The function \( f(x) = \frac{x^2 - 1}{x - 1} \) can be simplified:

\[
f(x) = \frac{(x - 1)(x + 1)}{x - 1}
\]

For \( x \neq 1 \), \( f(x) = x + 1 \). At \( x = 1 \), the function is undefined because the denominator becomes zero, which suggests a **removable discontinuity** at \( x = 1 \). If the function is redefined as \( f(1) = 2 \), it would be continuous.

### 1b) \( f(x) = x^2 - 2x + 3 \) at \( x = 1 \)

This is a polynomial function, which is continuous everywhere. Therefore, \( f(x) \) is continuous at \( x = 1 \).

### 2a) \( f(x) = \frac{2x + 3}{x - 2} \) at \( x = 2 \)

The function \( f(x) = \frac{2x + 3}{x - 2} \) is undefined at \( x = 2 \) because the denominator becomes zero. This suggests an **infinite discontinuity** at \( x = 2 \), where the function tends to infinity as \( x \) approaches 2.

### 2b) \( f(x) = \begin{cases} 
2x + 1 & \text{if } x < 1 \\
x^2 & \text{if } x \geq 1 
\end{cases} \) at \( x = 1 \)

We need to check the left-hand limit, right-hand limit, and the value of the function at \( x = 1 \):

- \( \lim\limits_{x \to 1^-} f(x) = 2(1) + 1 = 3 \)
- \( \lim\limits_{x \to 1^+} f(x) = 1^2 = 1 \)
- \( f(1) = 1^2 = 1 \)

Since the left-hand limit does not equal the right-hand limit, \( f(x) \) has a **jump discontinuity** at \( x = 1 \).

### 3a) \( f(x) = \frac{x^2 - x - 6}{x + 2} \) at \( x = -2 \)

The function \( f(x) = \frac{x^2 - x - 6}{x + 2} \) can be simplified by factoring the numerator:

\[
f(x) = \frac{(x - 3)(x + 2)}{x + 2}
\]

For \( x \neq -2 \), \( f(x) = x - 3 \). At \( x = -2 \), the function is undefined because the denominator becomes zero, indicating a **removable discontinuity** at \( x = -2 \).

### 3b) \( f(x) = \frac{x + 1}{x^2 - 4} \) at \( x = -2 \)

The function \( f(x) = \frac{x + 1}{x^2 - 4} \) can be factored as:

\[
f(x) = \frac{x + 1}{(x - 2)(x + 2)}
\]

At \( x = -2 \), the denominator becomes zero, so the function is undefined, which indicates an **infinite discontinuity** at \( x = -2 \).

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Would you like more detailed explanations, or do you have any other questions?

Here are 5 questions to consider:

1. How do you determine if a function has a removable discontinuity?
2. What is the difference between a jump discontinuity and an infinite discontinuity?
3. How can you redefine a function to make it continuous at a point with a removable discontinuity?
4. Why are polynomial functions always continuous?
5. How do you approach analyzing piecewise functions for continuity?

**Tip:** When dealing with rational functions, always check if the function can be simplified to identify potential removable discontinuities.

Determine if each function is continuous at the given x-values. If not continuous, classify each discontinuity.

Analyze functions to determine their continuity at specific x-values. Learn about types of discontinuities such as removable, infinite, and jump discontinuities. The problem covers key calculus concepts suitable for high school grades 11-12.

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