For the function `f(x) = x + 1/x, x in [1,3]` , the value of c for mean value therorem is

Options

A. 1
B. `sqrt(3)`
C. `2`
D. None of these

Explanation

To find the value of \( c \) for 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: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
In our case, \( f(x) \) is continuous and differentiable on \([1, 3]\), so we can apply MVT.

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

### Step 3: Calculate \( f(b) - f(a) \)
Now we find:
\[ f(b) - f(a) = f(3) - f(1) = \frac{10}{3} - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3} \]

### Step 4: Calculate \( b - a \)
\[ b - a = 3 - 1 = 2 \]

### Step 5: Apply the Mean Value Theorem formula
Now we substitute into the MVT formula:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3} \]

### Step 6: Find the derivative \( f'(x) \)
Next, we need to find the derivative of \( f(x) \):
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 7: Set the derivative equal to \(\frac{2}{3}\)
Now we set \( f'(c) \) equal to \(\frac{2}{3}\):
\[ 1 - \frac{1}{c^2} = \frac{2}{3} \]

### Step 8: Solve for \( c^2 \)
Rearranging gives:
\[ \frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3} \]
Thus,
\[ c^2 = 3 \]

### Step 9: Find \( c \)
Taking the square root gives:
\[ c = \sqrt{3} \quad \text{(since \( c \) must be in the interval (1, 3))} \]

### Final Answer
The value of \( c \) for the Mean Value Theorem is:
\[ \boxed{\sqrt{3}} \]