Find the absolute maximum and the absolute minimum value of the function given by :
`f(x)=sin^(2)x-cos x, x in [0, pi].`

Answer

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

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \sin^2 x - \cos x \]
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can rewrite the function as:
\[ f(x) = 1 - \cos^2 x - \cos x \]
This simplifies to:
\[ f(x) = -\cos^2 x - \cos x + 1 \]

### Step 2: Find the derivative
Next, we find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(-\cos^2 x - \cos x + 1) \]
Using the chain rule and the derivative of cosine, we get:
\[ f'(x) = -2\cos x(-\sin x) - (-\sin x) = 2\cos x \sin x + \sin x \]
Factoring out \( \sin x \):
\[ f'(x) = \sin x(2\cos x + 1) \]

### Step 3: Set the derivative to zero
To find critical points, we set the derivative equal to zero:
\[ \sin x(2\cos x + 1) = 0 \]
This gives us two cases:
1. \( \sin x = 0 \)
2. \( 2\cos x + 1 = 0 \)

**Case 1:** \( \sin x = 0 \)
In the interval \( [0, \pi] \), this occurs at:
\[ x = 0, \pi \]

**Case 2:** \( 2\cos x + 1 = 0 \)
Solving for \( \cos x \):
\[ \cos x = -\frac{1}{2} \]
In the interval \( [0, \pi] \), this occurs at:
\[ x = \frac{2\pi}{3} \]

### 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:

1. At \( x = 0 \):
\[ f(0) = \sin^2(0) - \cos(0) = 0 - 1 = -1 \]

2. At \( x = \pi \):
\[ f(\pi) = \sin^2(\pi) - \cos(\pi) = 0 - (-1) = 1 \]

3. At \( x = \frac{2\pi}{3} \):
\[ f\left(\frac{2\pi}{3}\right) = \sin^2\left(\frac{2\pi}{3}\right) - \cos\left(\frac{2\pi}{3}\right) \]
Calculating \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ f\left(\frac{2\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(-\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \]

### Step 5: Determine absolute maximum and minimum
Now we compare the values:
- \( f(0) = -1 \)
- \( f(\pi) = 1 \)
- \( f\left(\frac{2\pi}{3}\right) = \frac{5}{4} \)

The absolute maximum value is \( \frac{5}{4} \) at \( x = \frac{2\pi}{3} \) and the absolute minimum value is \( -1 \) at \( x = 0 \).

### Final Answer
- Absolute Maximum: \( \frac{5}{4} \) at \( x = \frac{2\pi}{3} \)
- Absolute Minimum: \( -1 \) at \( x = 0 \)

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