Examine the applicability of Rolle's Theorem for the function :
`f(x)=2+(x-1)^(2//3)` in the interval `0lexle2`.

Answer

To examine the applicability of Rolle's Theorem for the function \( f(x) = 2 + (x - 1)^{2/3} \) in the interval \( [0, 2] \), we will follow these steps:

### Step 1: Check the continuity of \( f(x) \) on the closed interval \( [0, 2] \)

The function \( f(x) = 2 + (x - 1)^{2/3} \) is a combination of a constant and a power function. The power function \( (x - 1)^{2/3} \) is continuous for all \( x \) since it is defined for all real numbers. Therefore, \( f(x) \) is continuous on the interval \( [0, 2] \).

### Step 2: Check the differentiability of \( f(x) \) on the open interval \( (0, 2) \)

To check differentiability, we need to find the derivative \( f'(x) \).

Using the power rule:
\[ f'(x) = \frac{d}{dx} \left( 2 + (x - 1)^{2/3} \right) = \frac{2}{3}(x - 1)^{-1/3} \cdot 1 = \frac{2}{3}(x - 1)^{-1/3} \]

Now, we need to check if \( f'(x) \) is defined for all \( x \) in the open interval \( (0, 2) \).

The derivative \( f'(x) \) is undefined at \( x = 1 \) because \( (x - 1)^{-1/3} \) becomes undefined (division by zero). Since \( x = 1 \) is in the interval \( (0, 2) \), the function is not differentiable at this point.

### Step 3: Check the values of \( f(a) \) and \( f(b) \)

Now we check the values of \( f \) at the endpoints of the interval:
- \( f(0) = 2 + (0 - 1)^{2/3} = 2 + 1 = 3 \)
- \( f(2) = 2 + (2 - 1)^{2/3} = 2 + 1 = 3 \)

Since \( f(0) = f(2) \), we have \( f(a) = f(b) \).

### Conclusion

Now we can summarize the conditions for Rolle's Theorem:
1. \( f(x) \) is continuous on \( [0, 2] \) - **True**
2. \( f(x) \) is differentiable on \( (0, 2) \) - **False** (not differentiable at \( x = 1 \))
3. \( f(0) = f(2) \) - **True**

Since the second condition is false, we conclude that **Rolle's Theorem is not applicable** for the function \( f(x) = 2 + (x - 1)^{2/3} \) on the interval \( [0, 2] \).