Question

18. Выясните, возрастает или убывает функция на указанном промежутке: 1) y= x>3; 1 (x-3)² 3) y=³x+1, x20; 3 2) y= 1 (x-2)8' x < 2 ; ,x​

Answer 1

Ответ:

To determine whether the function increases or decreases on the given interval, you need to find the derivative of each function and examine its sign.

1) For the function y = x^3, the derivative is dy/dx = 3x^2. Since the derivative is positive for all values of x, the function is increasing on the entire interval.

2) For the function y = (x-3)^2, the derivative is dy/dx = 2(x-3). The derivative is positive when x > 3 and negative when x < 3. Therefore, the function increases when x > 3 and decreases when x < 3.

3) For the function y = ³√x+1, the derivative is dy/dx = (1/3)(x+1)^(-2/3). Since the derivative is always positive for all values of x, the function is increasing on the entire interval.

4) The inequality x^2 > 3x+2 can be written as x^2 - 3x - 2 > 0. To find the values of x that make this inequality true, you can factorize the quadratic expression: (x-2)(x+1) > 0. For this inequality to be true, either both factors should be positive or both factors should be negative. This means x > 2 or x < -1. Therefore, the function y = x does not follow the inequality x < 2.

Therefore, the functions increase or decrease on the given interval as follows:

1) y = x^3: Increasing on the entire interval.

2) y = (x-3)^2: Increasing when x > 3 and decreasing when x < 3.

3) y = ³√x+1: Increasing on the entire interval.

4) y = x: This function does not follow the given inequality x < 2.