Which of the following functions satisfies all conditions 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 of the given functions satisfies all the conditions of Rolle's theorem in the specified intervals, we need to follow these steps:

### Step 1: Understand Rolle's Theorem
Rolle's theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Check Each Function
We will check each function provided in the options to see if they meet the conditions of Rolle's theorem.

#### Option 1: \( f(x) = x^{1/2} \) on \([-2, 3]\)
1. **Continuity**: The function \( f(x) = x^{1/2} \) is not defined for negative values of \( x \) (specifically at \( x = -2 \)). Therefore, it is not continuous on \([-2, 3]\).
2. **Conclusion**: This option does not satisfy Rolle's theorem.

#### Option 2: \( f(x) = \sin x \) on \([-5, \frac{\pi}{6}]\)
1. **Calculate \( f(-5) \)**:
\[ f(-5) = \sin(-5) \]
2. **Calculate \( f(\frac{\pi}{6}) \)**:
\[ f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} \]
3. **Check if \( f(-5) = f(\frac{\pi}{6}) \)**:
\[ \sin(-5) \neq \frac{1}{2} \]
4. **Conclusion**: This option does not satisfy Rolle's theorem.

#### Option 3: \( f(x) = e^x - x^2 \) on \([0, 4]\)
1. **Calculate \( f(0) \)**:
\[ f(0) = e^0 - 0^2 = 1 \]
2. **Calculate \( f(4) \)**:
\[ f(4) = e^4 - 4^2 = e^4 - 16 \]
3. **Check if \( f(0) = f(4) \)**:
\[ 1 \neq e^4 - 16 \]
4. **Conclusion**: This option does not satisfy Rolle's theorem.

#### Option 4: \( f(x) = \ln(x^2 + ab) \) on \([a, b]\)
1. **Calculate \( f(a) \)**:
\[ f(a) = \ln(a^2 + ab) \]
2. **Calculate \( f(b) \)**:
\[ f(b) = \ln(b^2 + ab) \]
3. **Check if \( f(a) = f(b) \)**:
\[ \ln(a^2 + ab) = \ln(b^2 + ab) \implies a^2 + ab = b^2 + ab \]
This simplifies to \( a^2 = b^2 \), which is true if \( a = b \) or \( a = -b \).
4. **Conclusion**: This option satisfies the condition \( f(a) = f(b) \).

### Final Conclusion
The function that satisfies all conditions of Rolle's theorem in the specified intervals is:

**Option 4: \( f(x) = \ln(x^2 + ab) \)** on \([a, b]\).