Verify Lagrange's Mean Value Theorem for the functions :
`f(x)=1/x` in the interval [-1, 2]

Answer

To verify Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = \frac{1}{x} \) on the interval \([-1, 2]\), we need to follow these steps:

### Step 1: Check the conditions of LMVT
Lagrange's 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:

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

Here, \( a = -1 \) and \( b = 2 \).

### Step 2: Determine continuity of \( f(x) \)
The function \( f(x) = \frac{1}{x} \) is continuous everywhere except at \( x = 0 \). Since the interval \([-1, 2]\) includes \( x = 0 \), the function is not continuous on this interval.

### Step 3: Determine differentiability of \( f(x) \)
Similarly, since \( f(x) \) is not defined at \( x = 0 \), it is also not differentiable at that point. Therefore, \( f(x) \) cannot be differentiable on the interval \((-1, 2)\) because it is not continuous at \( x = 0 \).

### Step 4: Conclusion
Since \( f(x) \) is not continuous on the closed interval \([-1, 2]\), the conditions for Lagrange's Mean Value Theorem are not satisfied. Thus, we conclude that LMVT is not applicable for the function \( f(x) = \frac{1}{x} \) on the interval \([-1, 2]\).

### Final Answer
Lagrange's Mean Value Theorem is not applicable for the function \( f(x) = \frac{1}{x} \) on the interval \([-1, 2]\) because the function is not continuous at \( x = 0 \).

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