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

Answer

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

### Step 1: Find the derivative of the function
We start by differentiating the function:
\[ f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ \cos x - \sin x = 0 \]
This simplifies to:
\[ \cos x = \sin x \]
Dividing both sides by \(\cos x\) (where \(\cos x \neq 0\)):
\[ 1 = \tan x \quad \Rightarrow \quad \tan x = 1 \]

### Step 3: Solve for \(x\)
The general solution for \(\tan x = 1\) is:
\[ x = n\pi + \frac{\pi}{4} \quad \text{where } n \text{ is an integer.} \]
In the interval \([0, \pi]\), the possible values are:
- For \(n = 0\): \(x = \frac{\pi}{4}\)
- For \(n = 1\): \(x = \frac{5\pi}{4}\) (not in the interval)

Thus, the only critical point in the interval \([0, \pi]\) is:
\[ x = \frac{\pi}{4} \]

### Step 4: Evaluate the function at the endpoints and critical points
Next, we evaluate \(f(x)\) at the endpoints \(x = 0\) and \(x = \pi\), and at the critical point \(x = \frac{\pi}{4}\):

1. **At \(x = 0\)**:
\[ f(0) = \sin(0) + \cos(0) = 0 + 1 = 1 \]

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

3. **At \(x = \frac{\pi}{4}\)**:
\[ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414 \]

### Step 5: Determine the absolute maximum and minimum
Now we compare the values:
- \(f(0) = 1\)
- \(f(\pi) = -1\)
- \(f\left(\frac{\pi}{4}\right) = \sqrt{2} \approx 1.414\)

From these values, we find:
- The **absolute maximum** value is \(\sqrt{2}\) at \(x = \frac{\pi}{4}\).
- The **absolute minimum** value is \(-1\) at \(x = \pi\).

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