a. Determine whether the Mean Value Theorem applies to the function $$ f(x)=x+\frac{1}{x} $$ on the interval $$ [-4,-1] $$. b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem. A. Choose the correct answer below. Bo, because the function is not continuous on the interval $$ [-4,-1] $$, and is not differentiable on the interval $$ (-4,-1) $$. C. Yes, because the function is continuous on the interval $$ [-4,-1] $$, but is not differentiable on the interval $$ (-4,-1) $$. D. No, because the function is is continuous on the interval $$ [-4,-1] $$ and differentiable on the interval $$ (-4,-1) $$. b. Select the correct choice below anderval $$ (-4,-1) $$, but is necessary, fill in the answer box to to complete your choice. A. The point(s) is/are $$ x=\square $$. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) $$ [-4,-1] $$. B. The Mean Value Theorem does not apply in this case.

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Let's analyze each part of the problem step by step.

### **a. Does the Mean Value Theorem Apply?**

**Mean Value Theorem (MVT) Requirements:**
1. The function must be **continuous** on the closed interval $$[a, b]$$.
2. The function must be **differentiable** on the open interval $$(a, b)$$.

**Given Function:**
$$ f(x) = x + \frac{1}{x} $$
**Interval:**
$$ [-4, -1] $$

1. **Continuity:**
   - The function $$ f(x) = x + \frac{1}{x} $$ is continuous everywhere except at $$ x = 0 $$ (where it has a vertical asymptote).
   - Since the interval $$[-4, -1]$$ does **not** include $$ x = 0 $$, $$ f(x) $$ is continuous on $$[-4, -1]$$.

2. **Differentiability:**
   - The derivative of $$ f(x) $$ is:
     $$ f'(x) = 1 - \frac{1}{x^2} $$
   - This derivative exists for all $$ x \neq 0 $$.
   - On the open interval $$(-4, -1)$$, $$ x \neq 0 $$, so $$ f(x) $$ is differentiable there.

**Conclusion for Part a:**
- **Yes**, the Mean Value Theorem applies because $$ f(x) $$ is continuous on $$[-4, -1]$$ and differentiable on $$(-4, -1)$$.

**Correct Choice:**
- It appears there might be a typo in the provided choices. However, based on the analysis, the correct statement should affirm that the MVT applies due to continuity and differentiability on the specified intervals.

### **b. Finding the Point Guaranteed by the Mean Value Theorem**

**Applying the Mean Value Theorem:**
The MVT states that there exists at least one point $$ c $$ in $$(-4, -1)$$ such that:
$$ f'(c) = \frac{f(-1) - f(-4)}{-1 - (-4)} = \frac{f(-1) - f(-4)}{3} $$

**Calculations:**
1. **Compute $$ f(-1) $$ and $$ f(-4) $$:**
   $$
   f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2
   $$
   $$
   f(-4) = -4 + \frac{1}{-4} = -4 - 0.25 = -4.25
   $$

2. **Average Rate of Change:**
   $$
   \frac{f(-1) - f(-4)}{3} = \frac{-2 - (-4.25)}{3} = \frac{2.25}{3} = 0.75
   $$

3. **Set the Derivative Equal to the Average Rate:**
   $$
   f'(c) = 1 - \frac{1}{c^2} = 0.75
   $$
   $$
   1 - \frac{1}{c^2} = 0.75 \implies \frac{1}{c^2} = 0.25 \implies c^2 = 4 \implies c = \pm 2
   $$
   
4. **Determine Valid $$ c $$ in $$(-4, -1)$$:**
   - $$ c = 2 $$ is **not** in $$(-4, -1)$$.
   - $$ c = -2 $$ **is** in $$(-4, -1)$$.

**Conclusion for Part b:**
- The point guaranteed by the Mean Value Theorem is $$ x = -2 $$.

**Correct Choice:**
- **A. The point(s) is/are $$ x = -2 $$.**

### **Final Answer:**

**a.** Yes, the Mean Value Theorem applies because $$ f(x) $$ is continuous on $$[-4, -1]$$ and differentiable on $$(-4, -1)$$.

**b.** The point guaranteed by the Mean Value Theorem is $$ x = -2 $$.
**a.** Yes, the Mean Value Theorem applies because $$ f(x) $$ is continuous on $$[-4, -1]$$ and differentiable on $$(-4, -1)$$. **b.** The point guaranteed by the Mean Value Theorem is $$ x = -2 $$.

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