Respuesta :

Answer:

range of the given function in Interval notation is [-1, 3]

Step-by-step explanation:

i) given function is y = 1 + 2sin(x - 11)

ii) maximum of sin function is 1

iii) therefore maximum of given function [tex]y_{max}[/tex] =  1 + (2[tex]\times[/tex] 1) = 3

iv) minimum of sin function is -1

v) therefore maximum of given function [tex]y_{max}[/tex] =  1 + (2[tex]\times[/tex] -1) = -1

vi) therefore the range of the given function in Interval notation is [-1,3]

Answer:

D. -1 to 3

Step-by-step explanation:

We have been given a function [tex]y=1+2sin(x-11 )[/tex]. We are asked to find the range of the function.

We know that since oscillates between [tex]-1[/tex] and 1, that is minimum value of sin is [tex]-1[/tex] and maximum is 1.    

The range of basic sin function is [tex]-1\leq \text{sin}(x-11)\leq 1[/tex].

Now we will multiply all sides by 2.

[tex]-1\cdot 2\leq 2\cdot \text{sin}(x-11)\leq 1\cdot 2[/tex]

[tex]-2\leq 2\text{sin}(x-11)\leq 2[/tex]

Now we will add 1 on all sides as:

[tex]-2+1\leq 2\text{sin}(x-11)+1\leq 2+1[/tex]

[tex]-1\leq 2\text{sin}(x-11)+1\leq 3[/tex]

[tex]-1\leq f(x) \leq 3[/tex]

Therefore, the range of the given function is [tex][-1,3][/tex] and option D is the correct choice.

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