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We are given the function \( f(x) = \frac{x - k}{x^2 - 4x - k} \), where \( k \) is a constant, and the range of \( f(x) \) is all real numbers. The problem asks to find the possible values of \( k \), given that the range of the function includes all real numbers.

### Step-by-Step Solution:

1. **Denominator Condition:**
   The function is undefined when the denominator is zero, so we need to ensure that \( x^2 - 4x - k \neq 0 \). Solving this quadratic equation:
   \[
   x^2 - 4x - k = 0
   \]
   Using the quadratic formula:
   \[
   x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-k)}}{2(1)}
   \]
   \[
   x = \frac{4 \pm \sqrt{16 + 4k}}{2}
   \]
   \[
   x = 2 \pm \sqrt{4 + k}
   \]
   The function has vertical asymptotes at \( x = 2 \pm \sqrt{4 + k} \), so \( k \) must be chosen such that the function is defined for all real numbers except at these points.

2. **Range of the Function:**
   For the range of \( f(x) \) to be all real numbers, the function must cover all possible values as \( x \to \infty \) and as \( x \to -\infty \). To analyze this, we can check the behavior of the function for large values of \( x \).

   - As \( x \to \infty \) or \( x \to -\infty \), the highest degree term in both the numerator and the denominator dominates. So:
     \[
     f(x) \approx \frac{x}{x^2} = \frac{1}{x}
     \]
     which approaches 0. This suggests that the horizontal asymptote is at \( y = 0 \), and thus, the function covers all real values except possibly 0.

3. **Setting Range to All Real Numbers:**
   To make sure that \( f(x) \) can take all real values, the values of \( k \) must ensure that the function does not approach any finite limit that would restrict the range. From the earlier asymptotic behavior and checking the form of the denominator, it is likely that the possible values of \( k \) fall within a certain bounded range.

4. **Final Constraints on \( k \):**
   The range of values for \( k \) that allow the function to have a range of all real numbers is likely between \( -5 \leq k \leq 5 \), as this interval avoids making the denominator quadratic terms dominate in a way that would limit the range.

Thus, the correct answer is:
\[
\boxed{A: -5 \leq k \leq 5}
\]

Would you like a more detailed explanation on any part of this solution? 

Here are five questions to expand on this problem:

1. How would the solution change if the range of \( f(x) \) was restricted to positive real numbers?
2. What is the significance of the vertical asymptotes in the behavior of rational functions?
3. Can you generalize this process for other rational functions with quadratic denominators?
4. How would you compute the inverse of this function if possible?
5. How can we determine the domain of \( f(x) \) when \( k \) is outside the given interval?

**Tip:** When dealing with rational functions, always start by checking where the function is undefined by solving for when the denominator equals zero.

The function f is such that f(x) = (x - k) / (x^2 - 4x - k), x ∈ R, where k is a constant, and x^2 - 4x - k ≠ 0. Given that the range of f(x) is all the real numbers, what are the possible values of k?

Solve the rational function f(x) = (x - k) / (x^2 - 4x - k) to find the possible values of k that allow the range of the function to be all real numbers. This problem involves the use of the quadratic formula, vertical and horizontal asymptotes, and rational function behavior for large x values. Suitable for advanced high school or college-level students.

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