Question

Найдите все значения x, при которых выполняется равенство f'(x) = 0, если f(x)=cos2x + x√3 и x Є [0;п] ?

Answer 1

F`(x)=-2sin2x+√3=0
sinx=√3/2
x=π/3+2πn U x=2π/3+2πk
0≤π/3+2πn≤π
0≤1+6n≤3
-1≤6n≤2
-1/6≤n≤1/3
n=0⇒x=π/3
0≤2π/3+2πk≤π
0≤2+6k≤3
-2≤6k≤1
-1/3≤k≤1/6
k=0⇒x=2π/3

Answer 2

Finding the Values of x for which f'(x) = 0

To find the values of x for which the equation f'(x) = 0 holds true, we need to first determine the derivative of the function f(x) = cos(2x) + x√3. Let's calculate the derivative of f(x) and then solve the equation f'(x) = 0.

The given function is f(x) = cos(2x) + x√3.

To find the derivative of f(x), we can use the chain rule and the power rule of differentiation. The derivative of cos(2x) is -2sin(2x) and the derivative of x√3 is √3.

Therefore, the derivative of f(x) is f'(x) = -2sin(2x) + √3.

Now, we need to solve the equation f'(x) = 0 to find the values of x.

Setting f'(x) = 0, we have -2sin(2x) + √3 = 0.

To solve this equation, we isolate sin(2x) by moving √3 to the other side:

-2sin(2x) = -√3.

Dividing both sides by -2, we get:

sin(2x) = √3/2.

Now, we need to find the values of x that satisfy this equation.

Using the unit circle and the properties of sine, we know that sin(2x) = √3/2 when 2x = π/3 or 2x = 2π/3.

Solving for x, we have:

2x = π/3, x = π/6.

2x = 2π/3, x = π/3.

Therefore, the values of x for which f'(x) = 0 are x = π/6 and x = π/3.

Since the given range for x is [0, п], we need to check if these values fall within the given range.

x = π/6 is approximately 0.5236, which falls within the range [0, п].

x = π/3 is approximately 1.0472, which also falls within the range [0, п].

Therefore, the values of x that satisfy the equation f'(x) = 0 and fall within the range [0, п] are x = π/6 and x = π/3.

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