Verify Rolle's Theorem for the function : f(x) = sin x + cos x in the interval `[0,2pi]`

Answer

To verify Rolle's Theorem for the function \( f(x) = \sin x + \cos x \) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Check the continuity and differentiability of \( f(x) \)

The function \( f(x) = \sin x + \cos x \) is composed of the sine and cosine functions, both of which are continuous and differentiable everywhere. Therefore, \( f(x) \) is continuous and differentiable on the interval \([0, 2\pi]\).

### Step 2: Evaluate \( f(a) \) and \( f(b) \)

We need to evaluate the function at the endpoints of the interval:
- Let \( a = 0 \) and \( b = 2\pi \).

Calculating \( f(0) \):
\[ f(0) = \sin(0) + \cos(0) = 0 + 1 = 1 \]

Calculating \( f(2\pi) \):
\[ f(2\pi) = \sin(2\pi) + \cos(2\pi) = 0 + 1 = 1 \]

### Step 3: Check if \( f(a) = f(b) \)

Since \( f(0) = 1 \) and \( f(2\pi) = 1 \), we have:
\[ f(0) = f(2\pi) \]

### Step 4: Find the derivative \( f'(x) \)

Now, we will find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]

### Step 5: Set the derivative equal to zero and solve for \( x \)

To find the critical points where \( f'(x) = 0 \):
\[ \cos x - \sin x = 0 \]
This implies:
\[ \cos x = \sin x \]
Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ 1 = \tan x \]
Thus, we find:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

For the interval \([0, 2\pi]\), the solutions are:
\[ x = \frac{\pi}{4} \quad \text{and} \quad x = \frac{5\pi}{4} \]

### Step 6: Verify that the critical points lie within the interval \([0, 2\pi]\)

Both \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \) are within the interval \([0, 2\pi]\).

### Conclusion

Since \( f(x) \) is continuous and differentiable on \([0, 2\pi]\), \( f(0) = f(2\pi) \), and we found points \( c \) in \((0, 2\pi)\) where \( f'(c) = 0\), we have verified that Rolle's Theorem holds for the function \( f(x) = \sin x + \cos x \) on the interval \([0, 2\pi]\).