`f(x) = sin^(4)x+cos^(4)x` in `[0,(pi)/(2)]`

Explanation

To solve the problem where \( f(x) = \sin^4 x + \cos^4 x \) in the interval \([0, \frac{\pi}{2}]\), we will follow these steps:

### Step 1: Check Continuity
First, we need to check if the function \( f(x) \) is continuous on the interval \([0, \frac{\pi}{2}]\).

**Solution:**
The functions \( \sin^4 x \) and \( \cos^4 x \) are both continuous functions because sine and cosine functions are continuous everywhere. The sum of continuous functions is also continuous. Therefore, \( f(x) = \sin^4 x + \cos^4 x \) is continuous on \([0, \frac{\pi}{2}]\).

### Step 2: Check Differentiability
Next, we will check if \( f(x) \) is differentiable on the interval \((0, \frac{\pi}{2})\).

**Solution:**
To find the derivative \( f'(x) \), we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^4 x) + \frac{d}{dx}(\cos^4 x) \]
Using the chain rule:
\[ f'(x) = 4\sin^3 x \cdot \cos x - 4\cos^3 x \cdot \sin x \]
This can be factored as:
\[ f'(x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x = 4\sin x \cos x (\sin^2 x - \cos^2 x) \]
Since \( \sin x \) and \( \cos x \) are continuous and differentiable in \((0, \frac{\pi}{2})\), \( f(x) \) is differentiable in this interval.

### Step 3: Apply Rolle's Theorem
Now we will check if the conditions of Rolle's Theorem are satisfied. We need to find \( f(0) \) and \( f\left(\frac{\pi}{2}\right) \).

**Solution:**
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 conditions of Rolle's Theorem are satisfied.

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

**Solution:**
Set the derivative equal to zero:
\[ 4\sin c \cos c (\sin^2 c - \cos^2 c) = 0 \]
This gives us two cases:
1. \( \sin c = 0 \) or \( \cos c = 0 \) (which do not lie in \((0, \frac{\pi}{2})\)).
2. \( \sin^2 c - \cos^2 c = 0 \) implies \( \sin^2 c = \cos^2 c \), leading to \( \tan^2 c = 1 \) or \( c = \frac{\pi}{4} \).

Thus, there exists at least one \( c \) in \((0, \frac{\pi}{2})\) such that \( f'(c) = 0 \).

### Final Conclusion
By applying Rolle's Theorem, we conclude that there exists at least one \( c \) in the interval \((0, \frac{\pi}{2})\) such that \( f'(c) = 0 \).

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