Find the minimum and maximum value of `sin^(4)x+cos^(2) x` where `x in R`

Answer

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

### Step 1: Rewrite the expression
We know that \( \sin^2 x + \cos^2 x = 1 \). Therefore, we can express \( \sin^2 x \) in terms of \( \cos^2 x \):
\[ \sin^2 x = 1 - \cos^2 x \]
Thus, we can rewrite \( \sin^4 x \) as:
\[ \sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2 \]

### Step 2: Expand the expression
Now, substituting this back into our original expression:
\[ \sin^4 x + \cos^2 x = (1 - \cos^2 x)^2 + \cos^2 x \]
Expanding \( (1 - \cos^2 x)^2 \):
\[ = 1 - 2\cos^2 x + \cos^4 x + \cos^2 x \]
Combining like terms:
\[ = 1 - 2\cos^2 x + \cos^4 x + \cos^2 x = 1 - \cos^2 x + \cos^4 x \]
Thus, we have:
\[ \sin^4 x + \cos^2 x = 1 - \cos^2 x + \cos^4 x \]

### Step 3: Substitute \( y = \cos^2 x \)
Let \( y = \cos^2 x \). Then, the expression becomes:
\[ 1 - y + y^2 \]
where \( y \) can take values in the interval \( [0, 1] \).

### Step 4: Find the derivative
To find the maximum and minimum values, we can differentiate the expression with respect to \( y \):
\[ f(y) = y^2 - y + 1 \]
Calculating the derivative:
\[ f'(y) = 2y - 1 \]
Setting the derivative to zero to find critical points:
\[ 2y - 1 = 0 \implies y = \frac{1}{2} \]

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

### Step 6: Determine minimum and maximum values
From our evaluations:
- The maximum value is \( 1 \) (at \( y = 0 \) and \( y = 1 \)).
- The minimum value is \( \frac{3}{4} \) (at \( y = \frac{1}{2} \)).

### Final Answer
Thus, the minimum value of \( \sin^4 x + \cos^2 x \) is \( \frac{3}{4} \) and the maximum value is \( 1 \).

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