Find the intervals in which the function 'f' given by :
`f(x)=sinx-cosx, 0lexle2pi`
is strictly increasing or striclty decreasing.

Answer

To determine the intervals in which the function \( f(x) = \sin x - \cos x \) is strictly increasing or strictly decreasing on the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x \]

### Step 2: Set the derivative equal to zero to find critical points
Next, we find the critical points by setting the derivative equal to zero:

\[ \cos x + \sin x = 0 \]

This can be rearranged to:

\[ \sin x = -\cos x \]

### Step 3: Solve for \( x \)
Dividing both sides by \( \cos x \) (where \( \cos x \neq 0 \)), we have:

\[ \tan x = -1 \]

The general solutions for \( \tan x = -1 \) are:

\[ x = \frac{3\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

Within the interval \( [0, 2\pi] \), the specific solutions are:

1. \( x = \frac{3\pi}{4} \)
2. \( x = \frac{7\pi}{4} \)

### Step 4: Determine the intervals
The critical points divide the interval \( [0, 2\pi] \) into three intervals:

1. \( [0, \frac{3\pi}{4}) \)
2. \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \)
3. \( (\frac{7\pi}{4}, 2\pi] \)

### Step 5: Test the sign of the derivative in each interval
We will test a point in each interval to determine whether \( f'(x) \) is positive (increasing) or negative (decreasing).

1. **Interval \( [0, \frac{3\pi}{4}) \):** Choose \( x = \frac{\pi}{4} \)

\[ f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} > 0 \]

Thus, \( f(x) \) is **increasing** on \( [0, \frac{3\pi}{4}) \).

2. **Interval \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \):** Choose \( x = \pi \)

\[ f'(\pi) = \cos(\pi) + \sin(\pi) = -1 + 0 = -1 < 0 \]

Thus, \( f(x) \) is **decreasing** on \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \).

3. **Interval \( (\frac{7\pi}{4}, 2\pi] \):** Choose \( x = \frac{3\pi}{2} \)

\[ f'(\frac{3\pi}{2}) = \cos(\frac{3\pi}{2}) + \sin(\frac{3\pi}{2}) = 0 - 1 = -1 < 0 \]

Thus, \( f(x) \) is **decreasing** on \( (\frac{7\pi}{4}, 2\pi] \).

### Conclusion
The function \( f(x) = \sin x - \cos x \) is:
- **Strictly increasing** on the interval \( [0, \frac{3\pi}{4}) \)
- **Strictly decreasing** on the intervals \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \) and \( (\frac{7\pi}{4}, 2\pi] \)