Rolle's Theorem for f (x) `=sin^4x+cos^4x` in `[0,(pi)/2]` is verified , then find c.

Answer

To verify Rolle's Theorem for the function \( f(x) = \sin^4 x + \cos^4 x \) in the interval \([0, \frac{\pi}{2}]\) and find the point \( c \), we will follow these steps:

### Step 1: Check the conditions of Rolle's Theorem
Rolle's Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one point \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Continuity**: The functions \( \sin x \) and \( \cos x \) are continuous everywhere, hence \( f(x) \) is continuous on \([0, \frac{\pi}{2}]\).
2. **Differentiability**: The functions \( \sin x \) and \( \cos x \) are differentiable everywhere, hence \( f(x) \) is differentiable on \((0, \frac{\pi}{2})\).
3. **Equal values at endpoints**:
- Calculate \( f(0) \):
\[ f(0) = \sin^4(0) + \cos^4(0) = 0 + 1 = 1 \]
- Calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \sin^4\left(\frac{\pi}{2}\right) + \cos^4\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]
- Since \( f(0) = f\left(\frac{\pi}{2}\right) = 1 \), the condition \( f(a) = f(b) \) is satisfied.

### Step 2: Differentiate the function
Next, we need to find the derivative \( f'(x) \):
\[ f(x) = \sin^4 x + \cos^4 x \]
Using the chain rule and product rule:
\[ f'(x) = 4\sin^3 x \cdot \cos x - 4\cos^3 x \cdot \sin x \]
Factoring out common terms:
\[ f'(x) = 4\sin x \cos x (\sin^2 x - \cos^2 x) \]

### Step 3: Set the derivative to zero
To find \( c \), we set \( f'(c) = 0 \):
\[ 4\sin c \cos c (\sin^2 c - \cos^2 c) = 0 \]
This gives us two cases:
1. \( \sin c = 0 \)
2. \( \cos c = 0 \)
3. \( \sin^2 c - \cos^2 c = 0 \)

### Step 4: Solve for \( c \)
1. From \( \sin c = 0 \), we have \( c = 0 \) (not in the open interval).
2. From \( \cos c = 0 \), we have \( c = \frac{\pi}{2} \) (not in the open interval).
3. From \( \sin^2 c - \cos^2 c = 0 \):
\[ \sin^2 c = \cos^2 c \implies \tan^2 c = 1 \implies \tan c = 1 \implies c = \frac{\pi}{4} \]

### Conclusion
Thus, the point \( c \) that satisfies Rolle's Theorem is:
\[ c = \frac{\pi}{4} \]