Question

For which interval(s) of the domain is the function invertible? Select ALL that apply. y = 3sin

A - [2(x- 3π /4 )]
B - [ π /4 , 3π /4 ]
C - [- π /4 , π /4 ]
D - [ π /2 ,π ]
E - [0, π /2 ]

Answer 1

The function y = 3sin(x) is invertible on intervals where it is one-to-one. The invertible intervals in the question are Option B [ π /4, 3π /4 ] and Option C [- π /4, π /4 ], as the sine function is either increasing or decreasing without repeating values.

Step-by-step explanation:

The function given is y = 3sin(x), and the question asks for intervals where this function is invertible. A function is invertible on an interval if it is one-to-one on that interval, meaning it passes the horizontal line test — each horizontal line will intersect the graph of the function at most once within that interval. Since the sine function has a period of 2π, we need to look for intervals where it is increasing or decreasing without repeating values.

• Option A cannot be correct because a sine function is not one-to-one over intervals that are greater than or equal to its period (2π).
• Option B: [ π /4, 3π /4 ] is an interval in which the sine function is decreasing and therefore is one-to-one, making it invertible.
• Option C: [- π /4, π /4 ] is an interval in which the sine function is increasing and therefore is one-to-one, making it invertible.
• Options D and E include points where the sine function reaches a maximum or minimum, at which the function stops increasing or decreasing, hence failing the horizontal line test.

Therefore, the intervals where the given function is invertible are Option B and Option C.