Question

Решите уравнение если f’(x)=0 , если f(x)=2 cos x+x

Answer 1

Производная равна -2sinx +1
f(0)=-2sin0+1=-2*0+1=1

Answer 2

To find the value of 'x' when the derivative of the function 'f(x)' is equal to zero, we need to solve the equation f'(x) = 0.

Given that f(x) = 2cos(x) + x, let's find the derivative of f(x) with respect to 'x' and set it equal to zero:

f(x) = 2cos(x) + x

To find f'(x): f'(x) = d/dx(2cos(x) + x) = -2sin(x) + 1.

Now, set f'(x) equal to zero and solve for 'x':

-2sin(x) + 1 = 0

Add 2sin(x) to both sides:

2sin(x) = 1

Now, divide both sides by 2:

sin(x) = 1/2

Now, to find the values of 'x' that satisfy this equation, we can use the inverse sine function (arcsin or sin^(-1)):

x = arcsin(1/2)

The arcsin(1/2) is equal to π/6 or 30 degrees (in degrees mode). However, remember that the sine function has periodicity, and thus it repeats itself. Therefore, there are more solutions.

The general solutions for sin(x) = 1/2 are:

x = π/6 + 2πn, where n is an integer.

So, the solutions for 'x' are:

1. x = π/6,
2. x = π/6 + 2π,
3. x = π/6 + 4π,
4. x = π/6 + 6π, ... and so on.

These solutions represent all the points where f'(x) = 0 for the given function f(x) = 2cos(x) + x.