Question

найдите наименьшее и наибольшее значение функциии f(x)=cosx-1/3cos3x на отрезке от 0 до П/2 . плиз помогите, очень прошу

Answer 1

 cos(0)-(1/3)*cos(0)=2/3

cos(pi/2)-(1/3)*cos(3*pi/2)=0

Минимальное значение 0, максимальное 2/3. Имхо

 

Но походу 00Саня00 прав. Я решил не верно. Тупанул, признаю

Answer 2

Finding the Minimum and Maximum Values of the Function f(x) = cos(x) - (1/3)cos(3x) on the Interval [0, π/2]

To find the minimum and maximum values of the function f(x) = cos(x) - (1/3)cos(3x) on the interval [0, π/2], we need to analyze the behavior of the function within this interval.

Let's start by finding the critical points of the function. Critical points occur where the derivative of the function is equal to zero or does not exist. We can then evaluate the function at these critical points and the endpoints of the interval to determine the minimum and maximum values.

Step 1: Finding the Critical Points

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

The derivative of f(x) is given by: f'(x) = -sin(x) + sin(3x)

Setting f'(x) equal to zero, we have: -sin(x) + sin(3x) = 0

Simplifying this equation, we get: sin(3x) = sin(x)

To solve this equation, we can use the trigonometric identity: sin(A) = sin(B) if and only if A = B + 2πn or A = π - B + 2πn, where n is an integer.

Applying this identity to our equation, we have two cases:

Case 1: sin(3x) = sin(x) In this case, we have: 3x = x + 2πn or 3x = π - x + 2πn

Simplifying these equations, we get: 2x = 2πn or 4x = π + 2πn

Solving for x, we have: x = πn or x = (π/4) + (π/2)n

Case 2: sin(3x) = -sin(x) In this case, we have: 3x = π - x + 2πn or 3x = -x + 2πn

Simplifying these equations, we get: 4x = π + 2πn or 4x = 2πn

Solving for x, we have: x = (π/4) + (π/2)n or x = (π/2)n

Therefore, the critical points of the function f(x) = cos(x) - (1/3)cos(3x) are: x = πn, (π/4) + (π/2)n, (π/2)n

Step 2: Evaluating the Function at the Critical Points and Endpoints

Now that we have the critical points, we can evaluate the function f(x) at these points and the endpoints of the interval [0, π/2].

The function f(x) = cos(x) - (1/3)cos(3x) is continuous on the interval [0, π/2], so the minimum and maximum values will occur at the critical points or the endpoints.

Let's evaluate the function at the critical points and endpoints:

f(0) = cos(0) - (1/3)cos(3(0)) = 1 - (1/3)(1) = 2/3 f(π/2) = cos(π/2) - (1/3)cos(3(π/2)) = 0 - (1/3)(-1) = 1/3

Now, let's evaluate the function at the critical points: f(π/4) = cos(π/4) - (1/3)cos(3(π/4)) = (√2/2) - (1/3)(√2/2) = (√2/2) - (√2/6) = (√2/3) f(π/2) = cos(π/2) - (1/3)cos(3(π/2)) = 0 - (1/3)(-1) = 1/3

Step 3: Determining the Minimum and Maximum Values

From the evaluations above, we have the following values:

f(0) = 2/3 f(π/4) = (√2/3) f(π/2) = 1/3

Therefore, the minimum value of the function f(x) = cos(x) - (1/3)cos(3x) on the interval [0, π/2] is 1/3, and the maximum value is 2/3.

Please note that the above calculations are based on the given function and the interval provided. If there are any