Among the following, the function (s) on which LMVT theorem is applicable in the indecatd intervals is/are

Options

A. `f(x)=x^((1)/(3))"in"[-1,1]`
B. `f(x)=x+1/x"in"[(1)/(2),3]`
C. `f(x)=(x-1)|(x-1)(x-2)|"in"[-1,1]`
D. `f(x)=e^(|(x-1)(x-3)|)"in"[1,3]`

Answer

To determine which functions satisfy the conditions of the Mean Value Theorem (MVT) in the indicated intervals, we will analyze each function step by step.

### Step 1: Understand the conditions of the Mean Value Theorem (MVT)
The MVT states that if a function \( f(x) \) is:
1. Continuous on the closed interval \([a, b]\)
2. Differentiable on the open interval \((a, b)\)

Then there exists at least one point \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

### Step 2: Analyze each function

**Option 1: \( f(x) = x^{1/3} \) on the interval \([-1, 1]\)**

- **Continuity**: The function \( f(x) = x^{1/3} \) is continuous on \([-1, 1]\).
- **Differentiability**: The derivative \( f'(x) = \frac{1}{3} x^{-2/3} \). At \( x = 0 \), \( f'(0) \) is undefined (as it leads to division by zero). Thus, \( f(x) \) is not differentiable at \( x = 0 \).

**Conclusion**: The MVT is not applicable for this function.

**Option 2: \( f(x) = x + \frac{1}{x} \) on the interval \([\frac{1}{2}, 3]\)**

- **Continuity**: The function is continuous on \([\frac{1}{2}, 3]\) since it is defined for all \( x \) in this interval.
- **Differentiability**: The derivative \( f'(x) = 1 - \frac{1}{x^2} \) is defined for all \( x \) in \((\frac{1}{2}, 3)\).

**Conclusion**: The MVT is applicable for this function.

**Option 3: \( f(x) = |x - 1| + x - 2 \) on the interval \([-1, 1]\)**

- **Continuity**: The function \( f(x) \) is continuous on \([-1, 1]\) since the absolute value function is continuous.
- **Differentiability**: The function \( f(x) \) has a sharp point at \( x = 1 \) (where the absolute value changes). However, in the open interval \((-1, 1)\), it does not touch this point, and thus it is differentiable.

**Conclusion**: The MVT is applicable for this function.

**Option 4: \( f(x) = e^{|x - 1|} \cdot (x - 3) \) on the interval \([1, 3]\)**

- **Continuity**: The function is continuous on \([1, 3]\) since both the exponential function and polynomial functions are continuous.
- **Differentiability**: The function is differentiable in the open interval \((1, 3)\) as it does not touch the sharp point at \( x = 1 \).

**Conclusion**: The MVT is applicable for this function.

### Final Conclusion
The functions for which the Mean Value Theorem is applicable in the indicated intervals are:
- Option 2: \( f(x) = x + \frac{1}{x} \)
- Option 3: \( f(x) = |x - 1| + x - 2 \)
- Option 4: \( f(x) = e^{|x - 1|} \cdot (x - 3) \)

### Summary of Results
- **Not Applicable**: Option 1
- **Applicable**: Options 2, 3, and 4