Verify Rolles theorem for function `f(x)=sin^4\ x+cos^4\ x` on `[0,\ pi/2]`

Explanation

To verify Rolle's Theorem for the function \( f(x) = \sin^4 x + \cos^4 x \) on the interval \([0, \frac{\pi}{2}]\), we need to follow these steps:

### Step 1: Check the continuity of \( f(x) \) on \([0, \frac{\pi}{2}]\)

The function \( f(x) = \sin^4 x + \cos^4 x \) is composed of the functions \( \sin^4 x \) and \( \cos^4 x \). Since both sine and cosine functions are continuous everywhere, their compositions and sums are also continuous. Therefore, \( f(x) \) is continuous on the interval \([0, \frac{\pi}{2}]\).

### Step 2: Check the differentiability of \( f(x) \) on \((0, \frac{\pi}{2})\)

Next, we need to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^4 x) + \frac{d}{dx}(\cos^4 x) \]
Using the chain rule, we find:
\[ f'(x) = 4\sin^3 x \cdot \cos x - 4\cos^3 x \cdot \sin x \]
This can be factored as:
\[ f'(x) = 4\sin x \cos x (\sin^2 x - \cos^2 x) \]
Since sine and cosine are differentiable everywhere, \( f(x) \) is differentiable on the open interval \((0, \frac{\pi}{2})\).

### Step 3: Check the values of \( f(x) \) at the endpoints

Now we evaluate \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = \sin^4(0) + \cos^4(0) = 0 + 1 = 1 \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \sin^4\left(\frac{\pi}{2}\right) + \cos^4\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]

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

### Step 4: Conclusion

Since all three conditions of Rolle's Theorem are satisfied (continuity on \([0, \frac{\pi}{2}]\), differentiability on \((0, \frac{\pi}{2})\), and \( f(0) = f\left(\frac{\pi}{2}\right) \)), there exists at least one \( c \) in \((0, \frac{\pi}{2})\) such that \( f'(c) = 0 \).

To find \( c \):
\[ f'(c) = 4\sin c \cos c (\sin^2 c - \cos^2 c) = 0 \]
This gives us two cases:
1. \( \sin c = 0 \) (not in \((0, \frac{\pi}{2})\))
2. \( \cos c = 0 \) (not in \((0, \frac{\pi}{2})\))
3. \( \sin^2 c - \cos^2 c = 0 \) implies \( \sin^2 c = \cos^2 c \) or \( \tan^2 c = 1 \), leading to \( c = \frac{\pi}{4} \).

Since \( \frac{\pi}{4} \) is in the interval \((0, \frac{\pi}{2})\), we conclude that Rolle's Theorem is verified.