Verify Rolle's Theorem for the functions :
`f(x)=cosx`, defined in the interval `[-pi/2,pi/2]`

Answer

To verify Rolle's Theorem for the function \( f(x) = \cos x \) defined on the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\), we will follow the steps outlined by the theorem.

### Step 1: Check Continuity
Rolle's Theorem states that the function must be continuous on the closed interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

**Solution:**
The function \( f(x) = \cos x \) is a cosine function, which is continuous everywhere on the real line. Therefore, it is continuous on the closed interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

### Step 2: Check Differentiability
Next, we need to check if the function is differentiable on the open interval \((- \frac{\pi}{2}, \frac{\pi}{2})\).

**Solution:**
The derivative of \( f(x) \) is given by:
\[ f'(x) = -\sin x \]
The sine function is also continuous and differentiable everywhere on the real line. Hence, \( f(x) \) is differentiable on the open interval \((- \frac{\pi}{2}, \frac{\pi}{2})\).

### Step 3: Check the Endpoints
We need to verify that \( f(a) = f(b) \) where \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \).

**Solution:**
Calculating the values at the endpoints:
\[ f(-\frac{\pi}{2}) = \cos(-\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \]
\[ f(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \]
Since \( f(-\frac{\pi}{2}) = f(\frac{\pi}{2}) = 0 \), the condition \( f(a) = f(b) \) is satisfied.

### Step 4: Find \( c \) such that \( f'(c) = 0 \)
According to Rolle's Theorem, there exists at least one \( c \) in the open interval \((- \frac{\pi}{2}, \frac{\pi}{2})\) such that \( f'(c) = 0 \).

**Solution:**
Setting the derivative to zero:
\[ f'(c) = -\sin c = 0 \]
This implies:
\[ \sin c = 0 \]
The solutions to \( \sin c = 0 \) within the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\) is:
\[ c = 0 \]
Since \( 0 \) lies within the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\), we have found our \( c \).

### Conclusion
Since all conditions of Rolle's Theorem are satisfied, we can conclude that:
\[ \text{Hence, Rolle's Theorem is verified.} \]

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