Verify Lagrange's Mean Value Theorem for the functions :
`f(x)=x^(1//3)` in the interval [-1, 1]

Answer

To verify Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = x^{1/3} \) in the interval \([-1, 1]\), we will follow these steps:

### Step 1: Check Continuity
The first step is to check if the function \( f(x) \) is continuous on the closed interval \([-1, 1]\).

The function \( f(x) = x^{1/3} \) is a root function, which is continuous for all real numbers. Therefore, it is continuous on the interval \([-1, 1]\).

**Hint:** A function is continuous on an interval if it does not have any breaks, jumps, or asymptotes within that interval.

### Step 2: Check Differentiability
Next, we need to check if the function is differentiable on the open interval \((-1, 1)\).

To find the derivative, we apply the power rule:
\[ f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3 \sqrt[3]{x^2}} \]

Now, we need to analyze the derivative \( f'(x) \):
- The derivative \( f'(x) \) is defined for all \( x \) except \( x = 0 \).

Since \( f'(x) \) is not defined at \( x = 0 \), the function is not differentiable at that point. Therefore, \( f(x) \) is not differentiable on the entire open interval \((-1, 1)\).

**Hint:** A function is differentiable at a point if its derivative exists at that point.

### Step 3: Apply Lagrange's Mean Value 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

In our case:
- \( a = -1 \)
- \( b = 1 \)
- \( f(-1) = (-1)^{1/3} = -1 \)
- \( f(1) = 1^{1/3} = 1 \)

Calculating the right-hand side:
\[ \frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - (-1)}{1 + 1} = \frac{2}{2} = 1 \]

Now, we need to check if there exists a \( c \) in \((-1, 1)\) such that \( f'(c) = 1 \). However, since \( f'(x) \) is not defined at \( x = 0 \), we cannot find such a \( c \).

### Conclusion
Since \( f(x) \) is not differentiable at \( x = 0 \), which lies in the interval \((-1, 1)\), the conditions of Lagrange's Mean Value Theorem are not satisfied. Therefore, we conclude that LMVT is not applicable for the function \( f(x) = x^{1/3} \) on the interval \([-1, 1]\).

**Final Answer:** Lagrange's Mean Value Theorem is not applicable for the function \( f(x) = x^{1/3} \) in the interval \([-1, 1]\).

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