F(x)=2 sinx+cos2x [0:п]

Предыдущий
вопрос Следующий
вопрос

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 =
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.