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 can follow these steps:

### Step 1: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function is continuous on a 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, \( f(x) = x + \frac{1}{x} \) is continuous and differentiable on the interval \([1, 3]\).

### Step 2: Calculate \( f(3) \) and \( f(1) \)
Now we need to find \( f(3) \) and \( f(1) \):
\[ f(3) = 3 + \frac{1}{3} = 3 + 0.333 = \frac{10}{3} \]
\[ f(1) = 1 + \frac{1}{1} = 1 + 1 = 2 \]

### Step 3: Apply the Mean Value Theorem formula
Now we can substitute these values into the MVT formula:
\[ f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{\frac{10}{3} - 2}{2} \]
Calculating \( f(3) - f(1) \):
\[ \frac{10}{3} - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3} \]
Now substituting back:
\[ f'(c) = \frac{\frac{4}{3}}{2} = \frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3} \]

### Step 4: Differentiate \( f(x) \)
Next, we need to find \( f'(x) \):
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 5: Set \( f'(c) \) equal to \( \frac{2}{3} \)
Now we set \( f'(c) \) equal to \( \frac{2}{3} \):
\[ 1 - \frac{1}{c^2} = \frac{2}{3} \]

### Step 6: Solve for \( c \)
Rearranging the equation:
\[ \frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3} \]
Taking the reciprocal:
\[ c^2 = 3 \]
Thus,
\[ c = \sqrt{3} \quad \text{(since } c \text{ must be positive in the interval [1, 3])} \]

### Conclusion
The value of \( c \) that satisfies the Mean Value Theorem for the function \( f(x) = x + \frac{1}{x} \) on the interval \([1, 3]\) is:
\[ c = \sqrt{3} \]