Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=cos^(2)x+sinx" in "[0, pi]`.

Answer

To find the absolute maximum and minimum values of the function \( f(x) = \cos^2 x + \sin x \) in the interval \( [0, \pi] \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by finding the derivative \( f'(x) \) of the function:
\[ f'(x) = \frac{d}{dx}(\cos^2 x + \sin x) \]
Using the chain rule and the derivative of sine and cosine, we get:
\[ f'(x) = 2\cos x(-\sin x) + \cos x = -2\cos x \sin x + \cos x \]
This can be simplified to:
\[ f'(x) = \cos x (1 - 2\sin x) \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:
\[ \cos x (1 - 2\sin x) = 0 \]
This gives us two cases:
1. \( \cos x = 0 \)
2. \( 1 - 2\sin x = 0 \)

### Step 3: Solve for critical points
**Case 1:** \( \cos x = 0 \)

The solutions for \( \cos x = 0 \) in the interval \( [0, \pi] \) is:
\[ x = \frac{\pi}{2} \]

**Case 2:** \( 1 - 2\sin x = 0 \)

Solving for \( \sin x \):
\[ \sin x = \frac{1}{2} \]
The solutions for \( \sin x = \frac{1}{2} \) in the interval \( [0, \pi] \) is:
\[ x = \frac{\pi}{6} \]

### Step 4: Evaluate the function at critical points and endpoints
Now, we evaluate \( f(x) \) at the critical points and the endpoints of the interval \( [0, \pi] \):
1. \( f(0) = \cos^2(0) + \sin(0) = 1 + 0 = 1 \)
2. \( f\left(\frac{\pi}{6}\right) = \cos^2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \)
3. \( f\left(\frac{\pi}{2}\right) = \cos^2\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1 \)
4. \( f(\pi) = \cos^2(\pi) + \sin(\pi) = 1 + 0 = 1 \)

### Step 5: Determine the absolute maximum and minimum
Now we compare the values:
- \( f(0) = 1 \)
- \( f\left(\frac{\pi}{6}\right) = \frac{5}{4} = 1.25 \)
- \( f\left(\frac{\pi}{2}\right) = 1 \)
- \( f(\pi) = 1 \)

The absolute maximum value is \( 1.25 \) at \( x = \frac{\pi}{6} \), and the absolute minimum value is \( 1 \) at \( x = 0, \frac{\pi}{2}, \) and \( \pi \).

### Final Answer
- **Absolute Maximum Value:** \( 1.25 \) at \( x = \frac{\pi}{6} \)
- **Absolute Minimum Value:** \( 1 \) at \( x = 0, \frac{\pi}{2}, \pi \)