Find `max^(m)&min^(m)` value of `sin^(2)theta+cos^(4)theta`

Answer

To find the maximum and minimum values of the function \( f(\theta) = \sin^2 \theta + \cos^4 \theta \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(\theta) = \sin^2 \theta + \cos^4 \theta \]

### Step 2: Use the Pythagorean identity
Recall that \( \sin^2 \theta + \cos^2 \theta = 1 \). We can express \( \cos^4 \theta \) in terms of \( \sin^2 \theta \) as follows:
\[ \cos^4 \theta = (\cos^2 \theta)^2 = (1 - \sin^2 \theta)^2 \]
Thus, we can rewrite \( f(\theta) \):
\[ f(\theta) = \sin^2 \theta + (1 - \sin^2 \theta)^2 \]

### Step 3: Let \( x = \sin^2 \theta \)
Let \( x = \sin^2 \theta \). Then, \( \cos^2 \theta = 1 - x \), and we can rewrite the function as:
\[ f(x) = x + (1 - x)^2 \]
Expanding this gives:
\[ f(x) = x + (1 - 2x + x^2) = x^2 - x + 1 \]

### Step 4: Find the derivative
To find the maximum and minimum values, we take the derivative of \( f(x) \):
\[ f'(x) = 2x - 1 \]
Setting the derivative equal to zero to find critical points:
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

### Step 5: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical point \( x = \frac{1}{2} \) and at the endpoints \( x = 0 \) and \( x = 1 \):
1. For \( x = 0 \):
\[ f(0) = 0 + (1 - 0)^2 = 1 \]
2. For \( x = 1 \):
\[ f(1) = 1 + (1 - 1)^2 = 1 \]
3. For \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \frac{1}{2} + \left(1 - \frac{1}{2}\right)^2 = \frac{1}{2} + \left(\frac{1}{2}\right)^2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]

### Step 6: Determine maximum and minimum values
From the evaluations:
- Maximum value of \( f(x) \) is \( 1 \) (at \( x = 0 \) and \( x = 1 \))
- Minimum value of \( f(x) \) is \( \frac{3}{4} \) (at \( x = \frac{1}{2} \))

### Final Result
Thus, the maximum and minimum values of the function \( \sin^2 \theta + \cos^4 \theta \) are:
\[ \text{Max} = 1, \quad \text{Min} = \frac{3}{4} \]