Verify the conditions of Mean Value Theorem in the following. In each case, find a point in the interval as stated by the Mean Value Theorem :
`f(x)=sinx-sin2x" on "[0,pi]`

Answer

To verify the conditions of the Mean Value Theorem (MVT) for the function \( f(x) = \sin x - \sin 2x \) on the interval \([0, \pi]\), we will follow these steps:

### Step 1: Check Continuity and Differentiability
The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
First, we need to check if \( f(x) \) is continuous and differentiable on the given interval.

- **Continuity**: The functions \( \sin x \) and \( \sin 2x \) are both continuous everywhere, hence \( f(x) \) is continuous on \([0, \pi]\).
- **Differentiability**: The functions \( \sin x \) and \( \sin 2x \) are also differentiable everywhere, thus \( f(x) \) is differentiable on \((0, \pi)\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Now we compute \( f(0) \) and \( f(\pi) \):
\[ f(0) = \sin(0) - \sin(2 \cdot 0) = 0 - 0 = 0 \]
\[ f(\pi) = \sin(\pi) - \sin(2\pi) = 0 - 0 = 0 \]

### Step 3: Apply the Mean Value Theorem
Now we can apply the MVT:
\[ f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} = \frac{0 - 0}{\pi} = 0 \]

### Step 4: Find \( f'(x) \)
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x - \sin 2x) = \cos x - 2\cos 2x \]

### Step 5: Set \( f'(x) = 0 \)
To find the points where \( f'(x) = 0 \):
\[ \cos x - 2\cos 2x = 0 \]
Using the double angle identity, \( \cos 2x = 2\cos^2 x - 1 \):
\[ \cos x - 2(2\cos^2 x - 1) = 0 \]
This simplifies to:
\[ \cos x - 4\cos^2 x + 2 = 0 \]
Rearranging gives:
\[ 4\cos^2 x - \cos x - 2 = 0 \]

### Step 6: Solve the Quadratic Equation
Let \( u = \cos x \). The equation becomes:
\[ 4u^2 - u - 2 = 0 \]
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{1 \pm \sqrt{1 + 32}}{8} = \frac{1 \pm \sqrt{33}}{8} \]

### Step 7: Find Valid Solutions
We need to check if \( u = \frac{1 + \sqrt{33}}{8} \) and \( u = \frac{1 - \sqrt{33}}{8} \) are within the range of \( \cos x \) (which must be between -1 and 1).

Calculating \( \frac{1 + \sqrt{33}}{8} \) and \( \frac{1 - \sqrt{33}}{8} \):
- \( \sqrt{33} \approx 5.744 \), so \( \frac{1 + \sqrt{33}}{8} \approx 0.843 \) (valid)
- \( \frac{1 - \sqrt{33}}{8} \approx -0.593 \) (valid)

### Step 8: Find \( c \)
Now we find \( c \) such that \( \cos c = \frac{1 + \sqrt{33}}{8} \) or \( \cos c = \frac{1 - \sqrt{33}}{8} \).

Thus, \( c \) can be calculated using:
\[ c = \cos^{-1}\left(\frac{1 + \sqrt{33}}{8}\right) \quad \text{or} \quad c = \cos^{-1}\left(\frac{1 - \sqrt{33}}{8}\right) \]
Both values of \( c \) will lie in the interval \((0, \pi)\).

### Conclusion
We have verified the conditions of the Mean Value Theorem and found points \( c \) in the interval \((0, \pi)\) where \( f'(c) = 0 \).