Verify Rolle's Theorem for the functions :
`f(x)=sin^(3)x+cos^(3)x` in the interval `[0,pi/2]`.

Answer

To verify Rolle's Theorem for the function \( f(x) = \sin^3 x + \cos^3 x \) in the interval \([0, \frac{\pi}{2}]\), we need to check the following conditions:

1. The function \( f(x) \) is continuous on the closed interval \([0, \frac{\pi}{2}]\).
2. The function \( f(x) \) is differentiable on the open interval \((0, \frac{\pi}{2})\).
3. The values of the function at the endpoints of the interval are equal, i.e., \( f(0) = f\left(\frac{\pi}{2}\right) \).
4. There exists at least one \( c \) in the open interval \((0, \frac{\pi}{2})\) such that \( f'(c) = 0 \).

### Step 1: Check Continuity
The functions \( \sin x \) and \( \cos x \) are continuous everywhere, and since \( f(x) \) is a combination of continuous functions, \( f(x) \) is continuous on the interval \([0, \frac{\pi}{2}]\).

### Step 2: Check Differentiability
Similarly, \( \sin x \) and \( \cos x \) are differentiable everywhere, so \( f(x) \) is differentiable on the open interval \((0, \frac{\pi}{2})\).

### Step 3: Check Endpoint Values
Now, we calculate the values of \( f(0) \) and \( f\left(\frac{\pi}{2}\right) \):
- \( f(0) = \sin^3(0) + \cos^3(0) = 0 + 1 = 1 \)
- \( f\left(\frac{\pi}{2}\right) = \sin^3\left(\frac{\pi}{2}\right) + \cos^3\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \)

Since \( f(0) = f\left(\frac{\pi}{2}\right) = 1 \), the third condition is satisfied.

### Step 4: Find \( c \) such that \( f'(c) = 0 \)
Now, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^3 x + \cos^3 x) \]
Using the chain rule:
\[ f'(x) = 3\sin^2 x \cdot \cos x - 3\cos^2 x \cdot \sin x \]
Factoring out \( 3\sin x \cos x \):
\[ f'(x) = 3\sin x \cos x (\sin x - \cos x) \]

Setting \( f'(x) = 0 \):
\[ 3\sin x \cos x (\sin x - \cos x) = 0 \]

This gives us two cases:
1. \( \sin x = 0 \) or \( \cos x = 0 \)
2. \( \sin x - \cos x = 0 \) which implies \( \tan x = 1 \)

From the first case:
- \( \sin x = 0 \) at \( x = 0 \) (not in the open interval).
- \( \cos x = 0 \) at \( x = \frac{\pi}{2} \) (not in the open interval).

From the second case:
- \( \tan x = 1 \) gives \( x = \frac{\pi}{4} \), which is in the open interval \((0, \frac{\pi}{2})\).

### Conclusion
Since all conditions of Rolle's Theorem are satisfied, we conclude that there exists at least one \( c \) in the interval \((0, \frac{\pi}{2})\) such that \( f'(c) = 0 \). Specifically, \( c = \frac{\pi}{4} \).