Which of the following function has / have a removable discontinuity at the indicated point ?

Options

A. `f(x) =(x^(2)-2x-8)/(x+2)` at x=-2
B. `f(x) =(x-7)/(|x-7|)` at x=7
C. `f(x) =(x^(3)+64)/(|x+4|)` at x=-4
D. `f(x) =(3-sqrt(x))/(9-x)` at x=9

Answer

To determine which of the given functions has a removable discontinuity at the indicated point, we will analyze each function step by step.

### Step 1: Analyze the first function
**Function:** \( f(x) = \frac{x^2 - 2x - 8}{x + 2} \) at \( x = -2 \)

1. **Factor the numerator:**
\[ x^2 - 2x - 8 = (x + 2)(x - 4) \]
So, we can rewrite \( f(x) \) as:
\[ f(x) = \frac{(x + 2)(x - 4)}{x + 2} \]
For \( x \neq -2 \), this simplifies to:
\[ f(x) = x - 4 \]

2. **Find the limit as \( x \) approaches -2:**
\[ \lim_{x \to -2} f(x) = -2 - 4 = -6 \]

3. **Check if the function can be defined at \( x = -2 \):**
If we define \( f(-2) = -6 \), the discontinuity at \( x = -2 \) can be removed.

### Conclusion for the first function:
The first function has a removable discontinuity at \( x = -2 \).

---

### Step 2: Analyze the second function
**Function:** \( f(x) = \frac{x - 7}{|x - 7|} \) at \( x = 7 \)

1. **Find the limit from the right:**
\[ \lim_{x \to 7^+} f(x) = \lim_{x \to 7^+} \frac{x - 7}{x - 7} = 1 \]

2. **Find the limit from the left:**
\[ \lim_{x \to 7^-} f(x) = \lim_{x \to 7^-} \frac{x - 7}{-(x - 7)} = -1 \]

3. **Check if the limits agree:**
Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

### Conclusion for the second function:
The second function does not have a removable discontinuity at \( x = 7 \).

---

### Step 3: Analyze the third function
**Function:** \( f(x) = \frac{x^3 + 64}{|x + 4|} \) at \( x = -4 \)

1. **Factor the numerator:**
\[ x^3 + 64 = (x + 4)(x^2 - 4x + 16) \]
So,
\[ f(x) = \frac{(x + 4)(x^2 - 4x + 16)}{|x + 4|} \]

2. **Find the limit from the right:**
\[ \lim_{x \to -4^+} f(x) = \lim_{x \to -4^+} (x^2 - 4x + 16) = 16 \]

3. **Find the limit from the left:**
\[ \lim_{x \to -4^-} f(x) = \lim_{x \to -4^-} -(x^2 - 4x + 16) = -16 \]

4. **Check if the limits agree:**
Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

### Conclusion for the third function:
The third function does not have a removable discontinuity at \( x = -4 \).

---

### Step 4: Analyze the fourth function
**Function:** \( f(x) = \frac{3 - \sqrt{x}}{9 - x} \) at \( x = 9 \)

1. **Rewrite the denominator:**
\[ 9 - x = (3 - \sqrt{x})(3 + \sqrt{x}) \]
So,
\[ f(x) = \frac{3 - \sqrt{x}}{(3 - \sqrt{x})(3 + \sqrt{x})} \]
For \( x \neq 9 \), this simplifies to:
\[ f(x) = \frac{1}{3 + \sqrt{x}} \]

2. **Find the limit as \( x \) approaches 9:**
\[ \lim_{x \to 9} f(x) = \frac{1}{3 + 3} = \frac{1}{6} \]

3. **Check if the function can be defined at \( x = 9 \):**
If we define \( f(9) = \frac{1}{6} \), the discontinuity at \( x = 9 \) can be removed.

### Conclusion for the fourth function:
The fourth function has a removable discontinuity at \( x = 9 \).

---

### Final Summary:
- The first function has a removable discontinuity at \( x = -2 \).
- The second function does not have a removable discontinuity at \( x = 7 \).
- The third function does not have a removable discontinuity at \( x = -4 \).
- The fourth function has a removable discontinuity at \( x = 9 \).