graph the function ( y = 2cosleft(\frac{1}{2}x\right) ). show at least two cycles. use the graph to…

graph the function ( y = 2cosleft(\frac{1}{2}x\right) ). show at least two cycles. use the graph to determine the domain and range of the function.\nuse the graphing tool to graph the equation. type pi to insert ( pi ) as needed.\nuse the graph to determine the domain of ( y = 2cosleft(\frac{1}{2}x\right) ).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)\nuse the graph to determine the range of ( y = 2cosleft(\frac{1}{2}x\right) ).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)

graph the function ( y = 2cosleft(\frac{1}{2}x\right) ). show at least two cycles. use the graph to determine the domain and range of the function.\nuse the graphing tool to graph the equation. type pi to insert ( pi ) as needed.\nuse the graph to determine the domain of ( y = 2cosleft(\frac{1}{2}x\right) ).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)\nuse the graph to determine the range of ( y = 2cosleft(\frac{1}{2}x\right) ).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)

Answer

Explanation:

Step1: Recall the domain of cosine function

The function (y = A\cos(Bx - C)+D) is a transformation of the basic cosine function (y=\cos(x)). The domain of (y = \cos(x)) is all real numbers, ((-\infty,\infty)). For the function (y = 2\cos(\frac{1}{2}x)), since there is no restriction on the value of (x) for which the cosine function is defined, the domain is based on the nature of the cosine function.

Step2: Recall the range of cosine function

The range of (y=\cos(x)) is ([- 1,1]). For the function (y = A\cos(Bx - C)+D), the range is given by ([D - |A|,D + |A|]). In the function (y = 2\cos(\frac{1}{2}x)), (A = 2), (B=\frac{1}{2}), (C = 0), (D = 0). Using the formula for the range of (y=A\cos(Bx - C)+D), we substitute (A = 2) and (D = 0) into ([D - |A|,D + |A|]).

Answer:

  • Domain: ((-\infty,\infty))
  • Range: ([-2,2])