The absolute maximum & minimum point of function f (x) = 2 cos 2x – cos 4x in [0, pi]
To find the absolute maximum and minimum points of the function f(x) = 2cos(2x) - cos(4x) on the interval [0, π], we need to find the critical points and endpoints and evaluate the function at these points.
1. Critical Points:
To find the critical points, we need to find the values of x where the derivative of the function is equal to 0 or does not exist.
Let's find the derivative of f(x):
f'(x) = -4sin(2x) + 4sin(4x)
Now, we set f'(x) = 0 and solve for x:
-4sin(2x) + 4sin(4x) = 0
Dividing both sides by 4:
-sin(2x) + sin(4x) = 0
Using the identity sin(4x) = 2sin(2x)cos(2x), we can rewrite the equation as:
-sin(2x) + 2sin(2x)cos(2x) = 0
Factoring out sin(2x):
sin(2x)(2cos(2x) - 1) = 0
Setting each factor equal to 0:
sin(2x) = 0 or 2cos(2x) - 1 = 0
1st factor: sin(2x) = 0
The solutions for sin(2x) = 0 in the interval [0, π] are x = 0 and x = π/2.
2nd factor: 2cos(2x) - 1 = 0
Adding 1 to both sides and dividing by 2:
cos(2x) = 1/2
The solutions for cos(2x) = 1/2 in the interval [0, π] are x = π/6 and x = 5π/6.
Therefore, the critical points are x = 0, x = π/6, x = π/2, and x = 5π/6.
2. Endpoints:
Since the interval [0, π] is closed, we need to evaluate the function at both endpoints.
When x = 0:
f(0) = 2cos(2(0)) - cos(4(0))
= 2cos(0) - cos(0)
= 2(1) - 1
= 1
When x = π:
f(π) = 2cos(2(π)) - cos(4(π))
= 2cos(2π) - cos(4π)
= 2(1) - cos(0)
= 1
Therefore, the function values at the endpoints are f(0) = 1 and f(π) = 1.
3. Evaluate f(x) at the critical points:
f(0) = 2cos(2(0)) - cos(4(0))
= 2(1) - cos(0)
= 2 - 1
= 1
f(π/6) = 2cos(2(π/6)) - cos(4(π/6))
= 2cos(π/3) - cos(2π/3)
= 2(1/2) - (-1/2)
= 1 + 1/2
= 3/2
f(π/2) = 2cos(2(π/2)) - cos(4(π/2))
= 2cos(π) - cos(2π)
= 2(-1) - 1
= -2 - 1
= -3
f(5π/6) = 2cos(2(5π/6)) - cos(4(5π/6))
= 2cos(5π/3) - cos(10π/3)
= 2(-1/2) - (1/2)
= -1 - 1/2
= -3/2
4. Comparing function values:
From our calculations, we have the following function values:
f(0) = 1
f(π/6) = 3/2
f(π/2) = -3
f(5π/6) = -3/2
f(π) = 1
The maximum value is 3/2, which occurs at x = π/6, and the minimum value is -3, which occurs at x = π/2.
Therefore, the absolute maximum point is (π/6, 3/2) and the absolute minimum point is (π/2, -3).