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Question
Which functions are continuous at x equals negative 2? Select all that apply.
(1 point)
Responses

f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared minus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row negative 4 if x equals negative 2 End Layout
Image with alt text: f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared minus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row negative 4 if x equals negative 2 End Layout

f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared minus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row 0 if x equals negative 2 End Layout
Image with alt text: f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared minus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row 0 if x equals negative 2 End Layout

f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared plus 4 x plus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row 0 if x equals negative 2 End Layout
Image with alt text: f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared plus 4 x plus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row 0 if x equals negative 2 End Layout

f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared plus 4 x plus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row 4 if x equals negative 2 End Layout
Image with alt text: f left-parenthesis x right-parenthesis equals enlarged left-brace Start Layout 1st row Start Fraction x squared plus 4 x plus 4 over x plus 2 End Fraction if x not equal to negative 2 2nd row 4 if x equals negative 2 End Layout
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Answer 1

To determine which functions are continuous at \( x = -2 \), we need to check the continuity condition:

A function \( f(x) \) is continuous at \( x = a \) if:
1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists.
3. \( \lim_{x \to a} f(x) = f(a) \).

We will analyze each given function:

1. **Function:**
\[
f(x) = \begin{cases}
\frac{x^2 - 4}{x + 2} & \text{if } x \neq -2 \\
-4 & \text{if } x = -2
\end{cases}
\]
- For \( x \neq -2 \), \( f(x) = \frac{x^2 - 4}{x + 2} = \frac{(x-2)(x+2)}{x+2} = x - 2 \) (for \( x \neq -2 \)).
- The limit as \( x \to -2 \) is \( -2 - 2 = -4 \).
- Since \( \lim_{x \to -2} f(x) = -4 \) and \( f(-2) = -4 \), this function is continuous at \( x = -2 \).

2. **Function:**
\[
f(x) = \begin{cases}
\frac{x^2 - 4}{x + 2} & \text{if } x \neq -2 \\
0 & \text{if } x = -2
\end{cases}
\]
- The limit as \( x \to -2 \) is \( -4 \).
- Since \( \lim_{x \to -2} f(x) = -4 \) and \( f(-2) = 0 \), this function is not continuous at \( x = -2 \).

3. **Function:**
\[
f(x) = \begin{cases}
\frac{x^2 + 4x + 4}{x + 2} & \text{if } x \neq -2 \\
0 & \text{if } x = -2
\end{cases}
\]
- For \( x \neq -2 \), \( f(x) = \frac{(x + 2)^2}{x + 2} = x + 2 \) (for \( x \neq -2 \)).
- The limit as \( x \to -2 \) is \( -2 + 2 = 0 \).
- Since \( \lim_{x \to -2} f(x) = 0 \) and \( f(-2) = 0 \), this function is continuous at \( x = -2 \).

4. **Function:**
\[
f(x) = \begin{cases}
\frac{x^2 + 4x + 4}{x + 2} & \text{if } x \neq -2 \\
4 & \text{if } x = -2
\end{cases}
\]
- The limit as \( x \to -2 \) is \( 0 \).
- Since \( \lim_{x \to -2} f(x) = 0 \) and \( f(-2) = 4 \), this function is not continuous at \( x = -2 \).

**Final Answer: The functions that are continuous at \( x = -2 \) are:**
- The first function
- The third function