Verify Lagrange 's mean value theorem for the function `f(x) = x+ (1)/(x) , 1 le x le 3 `

Answer

To verify 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: Check continuity and differentiability
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:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

1. **Continuity**: The function \( f(x) = x + \frac{1}{x} \) is a polynomial function and is continuous everywhere in its domain. Since \( x \) is positive in the interval \([1, 3]\), \( f(x) \) is continuous on \([1, 3]\).
2. **Differentiability**: The function \( f(x) \) is also differentiable everywhere in its domain, including the open interval \((1, 3)\).

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

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

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

### 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 \( c \) such that \( f'(c) = \frac{2}{3} \)
Next, we need to find \( f'(x) \):

\[ f'(x) = 1 - \frac{1}{x^2} \]

Now, we set \( f'(c) = \frac{2}{3} \):

\[ 1 - \frac{1}{c^2} = \frac{2}{3} \]

### Step 5: Solve for \( c \)
Rearranging the equation gives:

\[ \frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3} \]

Taking the reciprocal:

\[ c^2 = 3 \]

Thus,

\[ c = \sqrt{3} \]

### Step 6: Verify that \( c \) is in the interval \( (1, 3) \)
Since \( \sqrt{3} \approx 1.732 \), it is indeed in the interval \( (1, 3) \).

### Conclusion
We have found \( c = \sqrt{3} \) such that \( f'(c) = \frac{2}{3} \). Therefore, Lagrange's Mean Value Theorem is verified for the function \( f(x) = x + \frac{1}{x} \) on the interval \([1, 3]\).

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