For the function `f(x)=sin^(3)x-3sinx+4 AA x in [0, (pi)/(2)]`, which of the following is true?

Options

A. Greatest value of the function is `2pi`
B. Greatest value of the function is 4
C. Rolle's Theorem is applicable to `f(x)" in " x in [0, (pi)/(2)]`
D. LMVT is not applicable to f(x) in `x in [0, (pi)/(2)]`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^3 x - 3\sin x + 4 \) for \( x \) in the interval \( [0, \frac{\pi}{2}] \).

### Step 1: Evaluate the function at the endpoints
Let's first evaluate \( f(x) \) at the endpoints of the interval.

1. **At \( x = 0 \)**:
\[ f(0) = \sin^3(0) - 3\sin(0) + 4 = 0 - 0 + 4 = 4 \]

2. **At \( x = \frac{\pi}{2} \)**:
\[ f\left(\frac{\pi}{2}\right) = \sin^3\left(\frac{\pi}{2}\right) - 3\sin\left(\frac{\pi}{2}\right) + 4 = 1 - 3 + 4 = 2 \]

### Step 2: Determine the maximum value
From the evaluations:
- \( f(0) = 4 \)
- \( f\left(\frac{\pi}{2}\right) = 2 \)

The maximum value of \( f(x) \) on the interval \( [0, \frac{\pi}{2}] \) is \( 4 \) at \( x = 0 \).

### Step 3: Check the behavior of the function
Next, we need to check if the function is continuous and differentiable in the given interval.

1. **Continuity**: The function \( f(x) \) is a polynomial in terms of \( \sin x \) and is continuous everywhere, including the interval \( [0, \frac{\pi}{2}] \).

2. **Differentiability**: Since \( f(x) \) is composed of continuous functions, it is also differentiable in the open interval \( (0, \frac{\pi}{2}) \).

### Step 4: Apply Lagrange's Mean Value Theorem (LMVT)
Since \( f(x) \) is continuous on \( [0, \frac{\pi}{2}] \) and differentiable on \( (0, \frac{\pi}{2}) \), we can apply Lagrange's Mean Value Theorem. According to LMVT, there exists at least one \( c \) in \( (0, \frac{\pi}{2}) \) such that:
\[ f'(c) = \frac{f\left(\frac{\pi}{2}\right) - f(0)}{\frac{\pi}{2} - 0} = \frac{2 - 4}{\frac{\pi}{2}} = \frac{-2}{\frac{\pi}{2}} = -\frac{4}{\pi} \]

### Step 5: Check Rolle's Theorem conditions
Rolle's Theorem requires:
1. Continuity on \( [0, \frac{\pi}{2}] \) (which we have).
2. Differentiability on \( (0, \frac{\pi}{2}) \) (which we have).
3. \( f(0) = f\left(\frac{\pi}{2}\right) \) (which is not true since \( f(0) = 4 \) and \( f\left(\frac{\pi}{2}\right) = 2 \)).

Since the third condition is not satisfied, we cannot apply Rolle's Theorem.

### Conclusion
The maximum value of the function \( f(x) \) on the interval \( [0, \frac{\pi}{2}] \) is \( 4 \) at \( x = 0 \). The function satisfies the conditions for Lagrange's Mean Value Theorem but does not satisfy the conditions for Rolle's Theorem.

### Final Answer
- The maximum value of \( f(x) \) is \( 4 \) at \( x = 0 \).
- Lagrange's Mean Value Theorem applies.
- Rolle's Theorem does not apply.