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 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Here, \( a = 1 \) and \( b = 3 \). The function \( f(x) = x + \frac{1}{x} \) is continuous and differentiable on the interval \([1, 3]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Calculate \( f(1) \) and \( f(3) \):
\[ f(1) = 1 + \frac{1}{1} = 1 + 1 = 2 \]
\[ f(3) = 3 + \frac{1}{3} = 3 + 0.3333 \approx 3.3333 \]

### Step 3: Calculate the slope of the secant line
Now, we compute \( \frac{f(b) - f(a)}{b - a} \):
\[ \frac{f(3) - f(1)}{3 - 1} = \frac{3.3333 - 2}{2} = \frac{1.3333}{2} = 0.6667 \]

### Step 4: Find the derivative \( f'(x) \)
Next, we find the derivative of \( f(x) \):
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 5: Set up the equation for \( c \)
According to the MVT, we need to find \( c \) such that:
\[ f'(c) = \frac{f(3) - f(1)}{3 - 1} \]
This gives us:
\[ 1 - \frac{1}{c^2} = 0.6667 \]

### Step 6: Solve for \( c \)
Rearranging the equation:
\[ 1 - 0.6667 = \frac{1}{c^2} \]
\[ 0.3333 = \frac{1}{c^2} \]
Taking the reciprocal:
\[ c^2 = \frac{1}{0.3333} = 3 \]
Taking the square root:
\[ c = \sqrt{3} \]

### Conclusion
Thus, 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:
\[ c = \sqrt{3} \]