Question

Найти D(y)= E(y)= 1) y= 2x-1 2)y= 2x^2 3)y= -3x^3 4)y= (дробь) 2/x Помогите срочно!!!!!

Answer 1

1) y=2x-1 - линейная функция, поэтому D(y)=E(y)=(-∞;+∞)

2) y=2x² - квадратичная функция (график - парабола с вершиной в начале координат, ветви параболы направлены вверх, т.к. а=2>0), поэтому
  D(y)=(-∞;+∞)
  E(y)=[0;+∞)

3) y=-3x³ - кубическая парабола, поэтому D(y)=E(y)=(-∞;+∞)

4) y=2/x - гипербола с асимптотами х=0 и у=0, поэтому
   D(y)=E(y)=(-∞;0)U(0;+∞)

Answer 2

Finding the Value of D(y) and E(y)

To find the value of D(y) and E(y) for the given equations, let's go through each equation one by one:

1) y = 2x - 1 - D(y) = 2 (the coefficient of x) - E(y) = -1 (the constant term) 2) y = 2x^2 - D(y) = 4x (the derivative of 2x^2 with respect to x) - E(y) = 0 (there is no constant term) 3) y = -3x^3 - D(y) = -9x^2 (the derivative of -3x^3 with respect to x) - E(y) = 0 (there is no constant term) 4) y = 2/x - D(y) = -2/x^2 (the derivative of 2/x with respect to x) - E(y) = 0 (there is no constant term)

Therefore, the values of D(y) and E(y) for each equation are as follows:

1) D(y) = 2, E(y) = -1 2) D(y) = 4x, E(y) = 0 3) D(y) = -9x^2, E(y) = 0 4) D(y) = -2/x^2, E(y) = 0

Please let me know if you need any further assistance!