Question

Найдите точку максимума функции h(x)=sinx на отрезке [Pi/2; 3Pi/2]. С решением пожалуйста....

Answer 1

Ответ:

1. Тригонометрическая функция синус периодическая с периодом 2π. На промежутке [-π/2; π/2] возрастает, принимая значения от -1 до 1:

sin(-π/2) = -1;

sin0 = 0;

sin(π/2) = 1,

а на промежутке [π/2; 3π/2] убывает от 1 до -1:

sin(π/2) = 1,

sin(π) = 0;

sin(3π/2) = -1.

  2. Поскольку точки минимума 3π/2 и максимума π/2 принадлежат отрезку [-π/4; 3π/2], то наименьшее и наибольшее значения функции на заданном промежутке: -1 и 1.

  Ответ:

наименьшее значение: -1;

наибольшее значение: 1.

Answer 2

Finding the Maximum Point of the Function h(x) = sin(x) on the Interval [π/2, 3π/2]

To find the maximum point of the function h(x) = sin(x) on the interval [π/2, 3π/2], we need to determine the critical points and endpoints of the interval.

1. Critical Points: The critical points of a function occur where its derivative is equal to zero or undefined. Since the derivative of sin(x) is cos(x), we need to find where cos(x) = 0.

The cosine function is equal to zero at π/2 and 3π/2. Therefore, these points are potential critical points.

2. Endpoints: The interval [π/2, 3π/2] has two endpoints: π/2 and 3π/2.

3. Evaluating the Function at Critical Points and Endpoints: We need to evaluate the function h(x) = sin(x) at the critical points and endpoints to determine the maximum point.

- h(π/2) = sin(π/2) = 1 - h(3π/2) = sin(3π/2) = -1

4. Comparing the Values: Comparing the values of the function at the critical points and endpoints, we find that h(π/2) = 1 is the maximum value on the interval [π/2, 3π/2].

Therefore, the maximum point of the function h(x) = sin(x) on the interval [π/2, 3π/2] is (π/2, 1).

Note: The above solution is based on the properties of the sine function and the given interval.