Which of the following functions satisfies all contains of the Rolle's theorem in the invervals specified?

Options

A. `f(x)=x^((1)/(2)), x in [-2, 3]`
B. `f(x)=sinx,x in[-pi, (pi)/(6)]`
C. `f(x)=ln((x^(2)+ab)/(x(a+b))), x in [a, b], 0 lt a lt b`
D. `f(x)=e^(x^(2)-x), x in [0, 4]`

Answer

To determine which function satisfies all the conditions of Rolle's Theorem in the specified intervals, we need to follow these steps:

### Step 1: Understand the Conditions of Rolle's Theorem
Rolle's Theorem states that if a function \( f(x) \) is:
1. Continuous on the closed interval \([a, b]\),
2. Differentiable on the open interval \((a, b)\),
3. \( f(a) = f(b) \),

Then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Analyze Each Option
We will analyze each function provided in the options to check if they meet the conditions of Rolle's Theorem.

#### Option 1: \( f(x) = \sqrt{x} \)
- **Interval**: Let's say \( [a, b] = [-2, 3] \).
- **Continuity**: The function \( \sqrt{x} \) is not defined for negative values of \( x \). Therefore, it is not continuous on the interval \([-2, 3]\).
- **Conclusion**: This function does not satisfy the conditions of Rolle's Theorem.

#### Option 2: \( f(x) = \sin(x) \)
- **Interval**: Let's say \( [a, b] = [-\pi, \frac{\pi}{6}] \).
- **Continuity**: The sine function is continuous everywhere.
- **Differentiability**: The sine function is differentiable everywhere.
- **Values**: \( f(-\pi) = \sin(-\pi) = 0 \) and \( f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} \). Since \( f(-\pi) \neq f(\frac{\pi}{6}) \), this function does not satisfy the conditions of Rolle's Theorem.

#### Option 3: \( f(x) = \frac{x^2 + ab}{x(a + b)} \)
- **Interval**: Let's say \( [a, b] \) where \( a, b > 0 \).
- **Continuity**: The function is continuous for \( x > 0 \).
- **Differentiability**: The function is differentiable for \( x > 0 \).
- **Values**:
- \( f(a) = \frac{a^2 + ab}{a(a + b)} \)
- \( f(b) = \frac{b^2 + ab}{b(a + b)} \)

If we simplify \( f(a) \) and \( f(b) \), we find that \( f(a) = f(b) \) under certain conditions, thus satisfying \( f(a) = f(b) \).
- **Conclusion**: This function satisfies all conditions of Rolle's Theorem.

#### Option 4: \( f(x) = e^{x^2 - x} \)
- **Interval**: Let's say \( [0, 4] \).
- **Continuity**: The exponential function is continuous everywhere.
- **Differentiability**: The exponential function is differentiable everywhere.
- **Values**:
- \( f(0) = e^{0^2 - 0} = e^0 = 1 \)
- \( f(4) = e^{4^2 - 4} = e^{12} \neq 1 \)

Since \( f(0) \neq f(4) \), this function does not satisfy the conditions of Rolle's Theorem.

### Final Conclusion
The only function that satisfies all the conditions of Rolle's Theorem in the specified interval is **Option 3**.

---