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The problem in the image is asking for the value of \( g(a) \) for functions that have a removable discontinuity at \( x = a \). To solve this, we need to check for each function whether a removable discontinuity exists, and if it does, find the corresponding value of \( g(a) \) where the function would be continuous.

Let's go through each function:

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### (a) \( f(x) = \frac{x^2 - 2x - 8}{x + 2}, \ a = -2 \)

1. Factor the numerator: 
   \[
   x^2 - 2x - 8 = (x - 4)(x + 2)
   \]
2. The function becomes:
   \[
   f(x) = \frac{(x - 4)(x + 2)}{x + 2}
   \]
   For \( x \neq -2 \), we can cancel out \( (x + 2) \):
   \[
   f(x) = x - 4 \quad \text{for} \quad x \neq -2
   \]
3. To remove the discontinuity at \( x = -2 \), the value of \( g(-2) \) is the value of the simplified function \( x - 4 \) at \( x = -2 \):
   \[
   g(-2) = -2 - 4 = -6
   \]

---

### (b) \( f(x) = \frac{x - 7}{|x - 7|}, \ a = 7 \)

1. This function does **not** have a removable discontinuity at \( x = 7 \), as the absolute value function introduces a non-removable discontinuity (a jump discontinuity) at \( x = 7 \).
   
   Therefore, for this part, the answer is **n** (no removable discontinuity).

---

### (c) \( f(x) = \frac{x^3 + 64}{x + 4}, \ a = -4 \)

1. Factor the numerator using the sum of cubes:
   \[
   x^3 + 64 = (x + 4)(x^2 - 4x + 16)
   \]
2. The function becomes:
   \[
   f(x) = \frac{(x + 4)(x^2 - 4x + 16)}{x + 4}
   \]
   For \( x \neq -4 \), we can cancel \( (x + 4) \):
   \[
   f(x) = x^2 - 4x + 16 \quad \text{for} \quad x \neq -4
   \]
3. To remove the discontinuity at \( x = -4 \), the value of \( g(-4) \) is the value of the simplified function \( x^2 - 4x + 16 \) at \( x = -4 \):
   \[
   g(-4) = (-4)^2 - 4(-4) + 16 = 16 + 16 + 16 = 48
   \]

---

### (d) \( f(x) = \frac{3 - \sqrt{x}}{9 - x}, \ a = 9 \)

1. The function has a removable discontinuity if the numerator and denominator both go to zero at \( x = 9 \):
   \[
   f(9) = \frac{3 - \sqrt{9}}{9 - 9} = \frac{3 - 3}{0} = \frac{0}{0}
   \]
   This indicates a removable discontinuity at \( x = 9 \).
2. To remove the discontinuity, simplify the function near \( x = 9 \). Factor \( 9 - x \) as \( -(x - 9) \) and approximate for values close to 9. The limit of \( f(x) \) as \( x \) approaches 9 from both sides is:
   \[
   \lim_{x \to 9} \frac{3 - \sqrt{x}}{9 - x}
   \]
   Applying L'Hopital's rule:
   \[
   \lim_{x \to 9} \frac{\frac{-1}{2\sqrt{x}}}{-1} = \frac{1}{2\sqrt{9}} = \frac{1}{6}
   \]
   Thus, \( g(9) = \frac{1}{6} \).

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### Final Answers:

- (a) \( g(-2) = -6 \)
- (b) No removable discontinuity, answer is \( n \)
- (c) \( g(-4) = 48 \)
- (d) \( g(9) = \frac{1}{6} \)

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Would you like more details on any of these steps or further clarifications?

Here are 5 additional related questions:
1. How do you determine if a discontinuity is removable or not?
2. Can L'Hopital's rule be applied to every case of indeterminate forms?
3. What is the geometric interpretation of a removable discontinuity?
4. How does factoring help in finding removable discontinuities?
5. What is the difference between a removable and a jump discontinuity?

**Tip:** Always try simplifying a function before determining if a discontinuity is removable!

For the functions below that have a removable discontinuity at x = a, state the value of g(a), where g(x) agrees with f(x) for x ≠ a and is continuous everywhere.

Learn how to solve problems involving removable discontinuities in functions, including examples with factoring, limits, and L'Hopital's Rule. Suitable for high school students and college-level math learners.

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