Is LMV Theorem applicable to the function:
`f(x)=sinxsin2x` in the interval `[0,pi]`?

Answer

To determine if the Lagrange Mean Value Theorem (LMVT) is applicable to the function \( f(x) = \sin x \sin 2x \) on the interval \([0, \pi]\), we will follow these steps:

### Step 1: Check Continuity
The first condition of LMVT is that the function must be continuous on the closed interval \([a, b]\).

1. **Identify the function**: \( f(x) = \sin x \sin 2x \).
2. **Check the continuity of the components**: Both \( \sin x \) and \( \sin 2x \) are continuous functions for all \( x \).
3. **Product of continuous functions**: Since the product of continuous functions is also continuous, \( f(x) \) is continuous on \([0, \pi]\).

**Conclusion**: \( f(x) \) is continuous on \([0, \pi]\).

### Step 2: Check Differentiability
The second condition of LMVT is that the function must be differentiable on the open interval \((a, b)\).

1. **Differentiability of components**: Both \( \sin x \) and \( \sin 2x \) are differentiable for all \( x \).
2. **Product rule**: Therefore, \( f(x) = \sin x \sin 2x \) is differentiable on the open interval \((0, \pi)\).

**Conclusion**: \( f(x) \) is differentiable on \((0, \pi)\).

### Step 3: Apply Lagrange's Mean Value Theorem
The third condition states that there exists at least one \( c \) in the interval \((a, b)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

1. **Calculate \( f(0) \) and \( f(\pi) \)**:
- \( f(0) = \sin(0) \sin(0) = 0 \)
- \( f(\pi) = \sin(\pi) \sin(2\pi) = 0 \)

2. **Calculate \( f(b) - f(a) \)**:
- \( f(\pi) - f(0) = 0 - 0 = 0 \)

3. **Calculate \( b - a \)**:
- \( \pi - 0 = \pi \)

4. **Find the derivative \( f'(x) \)**:
- Using the product rule:
\[ f'(x) = \sin x \cdot \frac{d}{dx}(\sin 2x) + \sin 2x \cdot \frac{d}{dx}(\sin x) \]
- This gives:
\[ f'(x) = \sin x \cdot 2 \cos 2x + \sin 2x \cdot \cos x \]

5. **Set up the equation**:
- Since \( f(\pi) - f(0) = 0 \), we have:
\[ f'(c) = 0 \]

### Step 4: Solve for \( c \)
1. **Set the derivative equal to zero**:
\[ 2 \sin c \cos 2c + \sin 2c \cos c = 0 \]

2. **Factor out**:
- We can factor out \( \sin c \):
\[ \sin c (2 \cos 2c + \frac{\sin 2c}{\cos c}) = 0 \]
- This gives us two cases:
- \( \sin c = 0 \) (which gives \( c = 0 \) or \( c = \pi \), but these are not in the open interval)
- \( 2 \cos 2c + \frac{\sin 2c}{\cos c} = 0 \)

3. **Solve for \( c \)**:
- We find values of \( c \) that satisfy the equation in the interval \((0, \pi)\).

### Conclusion
Since we can find at least one \( c \) in the interval \((0, \pi)\) that satisfies the conditions of LMVT, we conclude that the Lagrange Mean Value Theorem is applicable to the function \( f(x) = \sin x \sin 2x \) on the interval \([0, \pi]\).

---