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To solve this problem, we need to find the global minimum value \( m \) and the global maximum value \( M \) of the function:

\[
f(x) = -x^3 - 6x^2 - 11x + 5
\]

on the interval \([-1, 1]\).

### Steps:
1. **Find the critical points:**
   Critical points occur where the derivative of the function equals zero or is undefined.

   - Compute the first derivative \( f'(x) \):
     \[
     f'(x) = \frac{d}{dx}(-x^3 - 6x^2 - 11x + 5) = -3x^2 - 12x - 11
     \]

   - Set \( f'(x) = 0 \) to find the critical points:
     \[
     -3x^2 - 12x - 11 = 0
     \]

     This is a quadratic equation, so we can solve it using the quadratic formula:
     \[
     x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
     \]
     where \( a = -3 \), \( b = -12 \), and \( c = -11 \).

2. **Evaluate the function at the critical points and endpoints:**
   After finding the critical points, evaluate \( f(x) \) at those points as well as the endpoints of the interval \( x = -1 \) and \( x = 1 \).

3. **Determine the global minimum and maximum:**
   Compare the function values to find the smallest and largest values, which will give the global minimum \( m \) and global maximum \( M \).

Let's solve this step-by-step. I'll start by calculating the critical points.The first derivative of the function is:

\[
f'(x) = -3x^2 - 12x - 11
\]

The critical points are:

\[
x = -2 - \frac{\sqrt{3}}{3} \quad \text{and} \quad x = -2 + \frac{\sqrt{3}}{3}
\]

Next, we will evaluate the function \( f(x) \) at these critical points and at the endpoints of the interval \([-1, 1]\). Since the critical points are outside the interval \([-1, 1]\), we only need to evaluate the function at the endpoints, \( x = -1 \) and \( x = 1 \). 

Let me calculate the values of \( f(x) \) at these points.The function values at the endpoints of the interval \([-1, 1]\) are:

- \( f(-1) = 11 \)
- \( f(1) = -13 \)

### Conclusion:
- The **global maximum** value \( M \) is 11 at \( x = -1 \).
- The **global minimum** value \( m \) is -13 at \( x = 1 \).

Therefore, the correct answer is:
\[
m = -13 \quad \text{and} \quad M = 11
\]

Would you like more details or have any questions about the process? Here are five related questions to deepen your understanding:

1. What are the conditions for a critical point to be a local maximum or minimum?
2. How can the second derivative test help in determining the nature of critical points?
3. Why do we need to consider the endpoints of a closed interval in optimization problems?
4. How is the quadratic formula derived, and when is it used in calculus?
5. What is the significance of concavity in finding extrema of functions?

**Tip:** Always check both the critical points and the endpoints when finding the global extrema of a function on a closed interval.

Consider the function f(x) = -x^3 - 6x^2 - 11x + 5. Find the global minimum value m and the global maximum value M of this function on the interval [-1, 1].

Learn how to find the global minimum and maximum of the function f(x) = -x^3 - 6x^2 - 11x + 5 over the interval [-1, 1]. This example covers key calculus concepts such as critical points and the use of the quadratic formula in optimization. Suitable for undergraduate calculus students.

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