The value of c in Lagrange's Mean Value theorem for the function f(x) `=x+(1)/(x)` in thhe interval [1,3] is

Options

A. 1
B. 2
C. `sqrt(3)`
D. `-sqrt(3)`

Answer

To find the value of \( c \) in Lagrange's Mean Value Theorem 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 Theorem
Lagrange's 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Here, \( f(x) = x + \frac{1}{x} \) is continuous and differentiable for all \( x > 0 \). Since our interval is \([1, 3]\), both conditions are satisfied.

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = 1 \) and \( b = 3 \).

Calculate \( f(1) \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

Calculate \( f(3) \):
\[ f(3) = 3 + \frac{1}{3} = 3 + 0.3333 = \frac{10}{3} \]

### Step 3: Calculate the Average Rate of Change
Now we compute the average rate of change of \( f \) over the interval \([1, 3]\):
\[ \frac{f(b) - f(a)}{b - a} = \frac{f(3) - f(1)}{3 - 1} = \frac{\frac{10}{3} - 2}{2} = \frac{\frac{10}{3} - \frac{6}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3} \]

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

### Step 5: Set \( f'(c) \) Equal to the Average Rate of Change
According to the Mean Value Theorem, we have:
\[ f'(c) = \frac{2}{3} \]
Substituting the expression for \( f'(x) \):
\[ 1 - \frac{1}{c^2} = \frac{2}{3} \]

### Step 6: Solve for \( c^2 \)
Rearranging the equation:
\[ -\frac{1}{c^2} = \frac{2}{3} - 1 \]
\[ -\frac{1}{c^2} = \frac{2}{3} - \frac{3}{3} = -\frac{1}{3} \]
Multiplying both sides by -1:
\[ \frac{1}{c^2} = \frac{1}{3} \]
Taking the reciprocal:
\[ c^2 = 3 \]
Thus,
\[ c = \sqrt{3} \]

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