Question

При каких значениях f'(x)=0, если f(x)=sin4xcosx-cos4xsinx+1.5x

Answer 1

По формуле синуса разности ф=синус 3х+1.5х берем производную 3косинуса 3х+1.5=0 косинус 3х=-1,5   косинус 3х=-0,5 3х=2/3п+2пк и 3х=-2/3п+2пк делим на 3  первый    х= 2п/9+2/3пк ,    второй   х=-2п/9+2/3пк

Answer 2

To find the critical points of the function f(x), we need to determine the values of x where the derivative f'(x) is equal to zero.

First, let's find the derivative f'(x) of the given function f(x):

f(x) = sin(4x)cos(x) - cos(4x)sin(x) + 1.5x

To find f'(x), we'll apply the product rule and the derivative of the linear term:

f'(x) = (d/dx)[sin(4x)cos(x)] - (d/dx)[cos(4x)sin(x)] + (d/dx)[1.5x]

Using the product rule: (d/dx)[sin(4x)cos(x)] = cos(4x)cos(x) - sin(4x)(-sin(x)) = cos(4x)cos(x) + sin(4x)sin(x)

(d/dx)[cos(4x)sin(x)] = -sin(4x)sin(x) - cos(4x)cos(x) = -sin(4x)sin(x) - cos(4x)cos(x)

(d/dx)[1.5x] = 1.5

Putting it all together:

f'(x) = cos(4x)cos(x) + sin(4x)sin(x) - sin(4x)sin(x) - cos(4x)cos(x) + 1.5

Now, simplify f'(x):

f'(x) = cos(4x)cos(x) - cos(4x)cos(x) + 1.5 f'(x) = 1.5

The derivative f'(x) is a constant, and there are no terms dependent on x. Therefore, there are no critical points in this case. The value of f'(x) is always 1.5, and it is never equal to zero.