Find the absolute maximum and minimum values of each of the following in the given intervals :
`y=2cos 2x-cos 4x, 0lexlepi.`

Answer

To find the absolute maximum and minimum values of the function \( y = 2 \cos(2x) - \cos(4x) \) on the interval \( [0, \pi] \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function with respect to \( x \):
\[ y = 2 \cos(2x) - \cos(4x) \]
Using the chain rule, we differentiate:
\[ \frac{dy}{dx} = -2 \sin(2x) \cdot 2 + 4 \sin(4x) = -4 \sin(2x) + 4 \sin(4x) \]
Thus, we have:
\[ \frac{dy}{dx} = 4 (\sin(4x) - \sin(2x)) \]

### Step 2: Set the derivative to zero to find critical points
To find critical points, we set the derivative equal to zero:
\[ 4 (\sin(4x) - \sin(2x)) = 0 \]
This simplifies to:
\[ \sin(4x) - \sin(2x) = 0 \]
Using the identity \( \sin(A) - \sin(B) = 0 \), we can set:
\[ \sin(4x) = \sin(2x) \]

### Step 3: Solve for \( x \)
The equation \( \sin(4x) = \sin(2x) \) gives us two cases:
1. \( 4x = 2x + n\pi \)
2. \( 4x = \pi - 2x + n\pi \)

From the first case:
\[ 2x = n\pi \implies x = \frac{n\pi}{2} \]
For \( n = 0, 1 \):
- \( x = 0 \)
- \( x = \frac{\pi}{2} \)

From the second case:
\[ 6x = \pi + n\pi \implies x = \frac{(n+1)\pi}{6} \]
For \( n = 0, 1 \):
- \( x = \frac{\pi}{6} \)
- \( x = \frac{7\pi}{6} \) (not in the interval)

Thus, the critical points in the interval \( [0, \pi] \) are:
- \( x = 0 \)
- \( x = \frac{\pi}{6} \)
- \( x = \frac{\pi}{2} \)

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( y \) at the critical points and endpoints:
1. At \( x = 0 \):
\[ y(0) = 2 \cos(0) - \cos(0) = 2 - 1 = 1 \]
2. At \( x = \frac{\pi}{6} \):
\[ y\left(\frac{\pi}{6}\right) = 2 \cos\left(\frac{\pi}{3}\right) - \cos\left(\frac{2\pi}{3}\right) = 2 \cdot \frac{1}{2} - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]
3. At \( x = \frac{\pi}{2} \):
\[ y\left(\frac{\pi}{2}\right) = 2 \cos(\pi) - \cos(2\pi) = 2 \cdot (-1) - 1 = -2 - 1 = -3 \]
4. At \( x = \pi \):
\[ y(\pi) = 2 \cos(2\pi) - \cos(4\pi) = 2 \cdot 1 - 1 = 2 - 1 = 1 \]

### Step 5: Compare values
Now we compare the values:
- \( y(0) = 1 \)
- \( y\left(\frac{\pi}{6}\right) = \frac{3}{2} \)
- \( y\left(\frac{\pi}{2}\right) = -3 \)
- \( y(\pi) = 1 \)

### Conclusion
The absolute maximum value is \( \frac{3}{2} \) and the absolute minimum value is \( -3 \).