Verify the conditions of Mean Value Theorem in the following. In each case, find a point in the interval as stated by the Mean Value Theorem :
`f(x)=x+1/x" on "[1,3]`

Answer

To verify the conditions of the Mean Value Theorem (MVT) for the function \( f(x) = x + \frac{1}{x} \) on the interval \([1, 3]\), we will follow these steps:

### Step 1: Check Continuity
The Mean Value Theorem states that a function must be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\).

1. **Continuity**: The function \( f(x) = x + \frac{1}{x} \) is composed of two parts: \( x \) and \( \frac{1}{x} \).
- The function \( x \) is continuous everywhere.
- The function \( \frac{1}{x} \) is continuous for all \( x \neq 0 \). Since our interval is \([1, 3]\), which does not include \( x = 0 \), \( f(x) \) is continuous on \([1, 3]\).

### Step 2: Check Differentiability
Next, we check if \( f(x) \) is differentiable on the open interval \((1, 3)\).

2. **Differentiability**: The function \( f(x) = x + \frac{1}{x} \) is differentiable everywhere in its domain. Since \( \frac{1}{x} \) is differentiable for \( x \neq 0 \), and since our interval \((1, 3)\) does not include \( x = 0 \), \( f(x) \) is differentiable on \((1, 3)\).

### Step 3: Apply the Mean Value Theorem
Since \( f(x) \) is continuous on \([1, 3]\) and differentiable on \((1, 3)\), we can apply the Mean Value Theorem.

3. **Mean Value Theorem Statement**: According to MVT, there exists at least one \( c \in (1, 3) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
where \( a = 1 \) and \( b = 3 \).

### Step 4: Calculate \( f(1) \) and \( f(3) \)
4. **Calculate \( f(1) \) and \( f(3) \)**:
\[ f(1) = 1 + \frac{1}{1} = 2 \]
\[ f(3) = 3 + \frac{1}{3} = 3 + 0.3333 = \frac{10}{3} \]

### Step 5: Calculate the Slope
5. **Calculate the slope**:
\[ \frac{f(3) - f(1)}{3 - 1} = \frac{\frac{10}{3} - 2}{2} = \frac{\frac{10}{3} - \frac{6}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{2}{3} \]

### Step 6: Find \( f'(x) \)
6. **Find the derivative \( f'(x) \)**:
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 7: Set \( f'(c) \) Equal to the Slope
7. **Set \( f'(c) = \frac{2}{3} \)**:
\[ 1 - \frac{1}{c^2} = \frac{2}{3} \]
Rearranging gives:
\[ \frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3} \]
Thus,
\[ c^2 = 3 \implies c = \sqrt{3} \]

### Step 8: Verify \( c \) is in the Interval
8. **Check if \( c \) is in the interval \((1, 3)\)**:
Since \( \sqrt{3} \approx 1.732 \), which lies in the interval \((1, 3)\), we have found our point.

### Conclusion
We have verified the conditions of the Mean Value Theorem and found that \( c = \sqrt{3} \) satisfies the theorem for the function \( f(x) = x + \frac{1}{x} \) on the interval \([1, 3]\).

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