To find the minimum and maximum values of the function F(x) = 2sin(x) + cos(2x) on the interval [0, π], we first need to find the critical points by taking the derivative of the function and setting it equal to zero.
F'(x) = 2cos(x) - 2sin(2x)
Setting F'(x) equal to zero:
2cos(x) - 2sin(2x) = 0
Now, we need to solve for x:
2cos(x) = 2sin(2x)cos(x) = sin(2x)cos(x) = 2sin(x)cos(x)1 = 2sin(x)sin(x) = 1/2
So, x = π/6.
Now, we need to check the endpoints of the interval [0, π] by evaluating F(0) and F(π):
F(0) = 2sin(0) + cos(0) = 0 + 1 = 1F(π) = 2sin(π) + cos(2π) = 0 + 1 = 1
Therefore, the minimum value of the function F(x) = 2sin(x) + cos(2x) on the interval [0, π] is 1 and the maximum value is 1.
To find the minimum and maximum values of the function F(x) = 2sin(x) + cos(2x) on the interval [0, π], we first need to find the critical points by taking the derivative of the function and setting it equal to zero.
F'(x) = 2cos(x) - 2sin(2x)
Setting F'(x) equal to zero:
2cos(x) - 2sin(2x) = 0
Now, we need to solve for x:
2cos(x) = 2sin(2x)
cos(x) = sin(2x)
cos(x) = 2sin(x)cos(x)
1 = 2sin(x)
sin(x) = 1/2
So, x = π/6.
Now, we need to check the endpoints of the interval [0, π] by evaluating F(0) and F(π):
F(0) = 2sin(0) + cos(0) = 0 + 1 = 1
F(π) = 2sin(π) + cos(2π) = 0 + 1 = 1
Therefore, the minimum value of the function F(x) = 2sin(x) + cos(2x) on the interval [0, π] is 1 and the maximum value is 1.