Question

Укажите D(y) E(y) 1) y= -|2cos(0.5x)| +4 2) y= 3sin (2x - Pi/3) - 2

Answer 1

1)D(у)=(-∞;+∞)

0≤|2cos(0.5x)| ≤2

-2≤|2cos(0.5x)| ≤0

2≤-|2cos(0.5x)| +4≤ 4

Е(у)=[2;4]

2)  y= 3sin (2x - π/3) - 2; D(у)=(-∞;+∞)

-3≤3sin (2x - π/3)≤3

-2-3≤3sin (2x - π/3) - 2≤3-2

-5≤3sin (2x - π/3) - 2≤1

Е(у)=[-5;1]

-2-3≤3sin (2x - π/3) - 2≤3-2

-2-3≤3sin (2x - π/3) - 2≤3-2

Answer 2

To find the domain and the range of the given functions, we need to determine the possible values for 'y' and 'x'.

1. y = -|2cos(0.5x)| + 4

Domain (D(y)): The domain of a function refers to the set of all possible values for the independent variable 'x' for which the function is defined. In this case, since cosine is defined for all real numbers and the absolute value of a real number is also real, there are no restrictions on 'x' that make the expression undefined. Hence, the domain is all real numbers:

D(y) = (-∞, ∞)

Range (E(y)): The range of a function refers to the set of all possible values for the dependent variable 'y'. To find the range, we need to consider the possible values of the expression -|2cos(0.5x)| + 4.

Since the cosine function oscillates between -1 and 1, the expression -|2cos(0.5x)| can take values from -2 to 2. Adding 4 to these values will shift the range upward:

E(y) = (-2 + 4, 2 + 4) = (2, 6)

So, the range of the function is:

E(y) = (2, 6)

2. y = 3sin(2x - π/3) - 2

Domain (D(y)): The domain is still all real numbers because the sine function is defined for any real value of its argument (2x - π/3).

D(y) = (-∞, ∞)

Range (E(y)): The range of the sine function is between -1 and 1. Multiplying by 3 will change the amplitude to 3, and subtracting 2 will shift the range downward.

E(y) = (3 * (-1) - 2, 3 * 1 - 2) = (-5, 1)

So, the range of the function is:

E(y) = (-5, 1)

To summarize:

1. For the function y = -|2cos(0.5x)| + 4: Domain (D(y)) = (-∞, ∞) Range (E(y)) = (2, 6)

2. For the function y = 3sin(2x - π/3) - 2: Domain (D(y)) = (-∞, ∞) Range (E(y)) = (-5, 1)