The minimum value of `f(x)-sin^(4)x+cos^(4)x,0lexle(pi)/(2)` is

Options

A. `(1)/(2sqrt2)`
B. `1/4`
C. `-1/2`
D. `1/2`

Answer

To find the minimum value of the function \( f(x) = \sin^4 x + \cos^4 x \) for \( 0 \leq x \leq \frac{\pi}{2} \), we can follow these steps:

### Step 1: Rewrite the function
We can use the identity \( \sin^4 x + \cos^4 x \) to simplify our calculations.
Using the identity:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]

### Step 2: Use the double angle identity
We can express \( \sin^2 x \cos^2 x \) using the double angle identity:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
Thus, we can rewrite our function as:
\[ f(x) = 1 - \frac{1}{2} \sin^2(2x) \]

### Step 3: Analyze the function
The function \( \sin^2(2x) \) takes values from 0 to 1. Therefore, \( f(x) \) will take values from:
\[ f(x) = 1 - \frac{1}{2} \cdot 1 = \frac{1}{2} \quad \text{(maximum of } \sin^2(2x) \text{)} \]
to
\[ f(x) = 1 - \frac{1}{2} \cdot 0 = 1 \quad \text{(minimum of } \sin^2(2x) \text{)} \]

### Step 4: Find the minimum value
Thus, the minimum value of \( f(x) \) occurs when \( \sin^2(2x) \) is maximized, which is at \( \sin^2(2x) = 1 \). Therefore, the minimum value of \( f(x) \) is:
\[ \frac{1}{2} \]

### Conclusion
The minimum value of \( f(x) = \sin^4 x + \cos^4 x \) for \( 0 \leq x \leq \frac{\pi}{2} \) is:
\[ \boxed{\frac{1}{2}} \]