Verify Lagrange's Mean Value Theorem for the functions :
`f(x)=(x-1)^(2//3)` in the interval [1, 2].

Answer

To verify Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = (x-1)^{2/3} \) on the interval \([1, 2]\), we will follow these steps:

### Step 1: Check if the function is continuous on the interval \([1, 2]\)

The function \( f(x) = (x-1)^{2/3} \) is a composition of continuous functions (the power function and the polynomial function). Since the base \( (x-1) \) is non-negative for \( x \in [1, 2] \), \( f(x) \) is continuous on this interval.

**Hint:** A function is continuous on an interval if it is defined at every point in that interval and does not have any jumps or breaks.

### Step 2: Check if the function is differentiable on the interval \((1, 2)\)

To check differentiability, we need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}[(x-1)^{2/3}] = \frac{2}{3}(x-1)^{-1/3} \]
The derivative \( f'(x) \) is defined for \( x > 1 \) (i.e., on the interval \((1, 2)\)). At \( x = 1 \), the derivative is not defined because it involves division by zero. However, since we are only concerned with the open interval \((1, 2)\), \( f(x) \) is differentiable on this interval.

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

### Step 3: Calculate \( f(1) \) and \( f(2) \)

Now we compute the values of the function at the endpoints of the interval:
\[ f(1) = (1-1)^{2/3} = 0 \]
\[ f(2) = (2-1)^{2/3} = 1^{2/3} = 1 \]

### Step 4: Apply Lagrange's Mean Value Theorem

According to LMVT, there exists at least one \( c \) in the interval \((1, 2)\) such that:
\[ f'(c) = \frac{f(2) - f(1)}{2 - 1} \]
Calculating the right-hand side:
\[ f'(c) = \frac{1 - 0}{1} = 1 \]

### Step 5: Set the derivative equal to 1 and solve for \( c \)

We need to find \( c \) such that:
\[ f'(c) = \frac{2}{3}(c-1)^{-1/3} = 1 \]
Solving for \( c \):
\[ \frac{2}{3}(c-1)^{-1/3} = 1 \]
Multiplying both sides by \( (c-1)^{1/3} \):
\[ \frac{2}{3} = (c-1)^{1/3} \]
Cubing both sides:
\[ \left(\frac{2}{3}\right)^3 = c - 1 \]
\[ \frac{8}{27} = c - 1 \]
Thus,
\[ c = 1 + \frac{8}{27} = \frac{27 + 8}{27} = \frac{35}{27} \]

### Step 6: Verify that \( c \) is in the interval \((1, 2)\)

Now, we check if \( c = \frac{35}{27} \) lies within the interval \((1, 2)\):
\[ 1 < \frac{35}{27} < 2 \]
Since \( \frac{35}{27} \approx 1.296 \), it is indeed in the interval \((1, 2)\).

### Conclusion

Thus, we have verified Lagrange's Mean Value Theorem for the function \( f(x) = (x-1)^{2/3} \) on the interval \([1, 2]\).

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