Find the interval(s) in which `f(x)= cos(2x+(pi)/(4)), 0 le x le pi` is increasing.

Answer

To find the interval(s) in which the function \( f(x) = \cos(2x + \frac{\pi}{4}) \) is increasing for \( 0 \leq x \leq \pi \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx} \left( \cos(2x + \frac{\pi}{4}) \right) \]

Using the chain rule, we have:

\[ f'(x) = -\sin(2x + \frac{\pi}{4}) \cdot \frac{d}{dx}(2x + \frac{\pi}{4}) = -\sin(2x + \frac{\pi}{4}) \cdot 2 \]

Thus,

\[ f'(x) = -2\sin(2x + \frac{\pi}{4}) \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we need to set the derivative greater than zero:

\[ f'(x) > 0 \implies -2\sin(2x + \frac{\pi}{4}) > 0 \]

This simplifies to:

\[ \sin(2x + \frac{\pi}{4}) < 0 \]

### Step 3: Determine where the sine function is negative
The sine function is negative in the intervals:

\[ \left( \pi, 2\pi \right) \]

Thus, we need to solve the inequality:

\[ \pi < 2x + \frac{\pi}{4} < 2\pi \]

### Step 4: Solve the inequalities
We will solve the two inequalities separately.

1. For the left inequality:

\[ \pi < 2x + \frac{\pi}{4} \]

Subtracting \( \frac{\pi}{4} \) from both sides:

\[ \pi - \frac{\pi}{4} < 2x \]

This simplifies to:

\[ \frac{3\pi}{4} < 2x \]

Dividing by 2:

\[ \frac{3\pi}{8} < x \]

2. For the right inequality:

\[ 2x + \frac{\pi}{4} < 2\pi \]

Subtracting \( \frac{\pi}{4} \) from both sides:

\[ 2x < 2\pi - \frac{\pi}{4} \]

This simplifies to:

\[ 2x < \frac{8\pi}{4} - \frac{\pi}{4} = \frac{7\pi}{4} \]

Dividing by 2:

\[ x < \frac{7\pi}{8} \]

### Step 5: Combine the results
From the two inequalities, we find:

\[ \frac{3\pi}{8} < x < \frac{7\pi}{8} \]

### Conclusion
Thus, the function \( f(x) = \cos(2x + \frac{\pi}{4}) \) is increasing in the interval:

\[ \left( \frac{3\pi}{8}, \frac{7\pi}{8} \right) \]