Final answer:
The sine function f(x) = sin(x) is invertible on the domains [0, π] and [-π/2, π/2], as each exhibits a one-to-one correspondence between x and f(x) values within these ranges. The domain [2π, 3π] does not maintain this one-to-one relationship.
Step-by-step explanation:
The function f(x) = sin(x) is considered invertible when it has a one-to-one correspondence within a certain domain, meaning each input x has exactly one output f(x) and vice versa. For sine functions, which oscillate between -1 and 1 and repeat every 2π radians, certain restricted domains are necessary to maintain one-to-oneness and therefore invertibility.
Domain a. [0, π] encompasses one full sine wave cycle, ranging from the sine function's maximum at π/2 back down to zero. This domain is invertible because within this interval, each value of sin(x) corresponds to exactly one x-value. Similarly, domain b. [-π/2, π/2] is also invertible. It starts at the sine function's midpoint (sine equals zero at -π/2), ascends to the maximum at x=0, and descends back to zero at x=π/2.
However, domain d. [2π, 3π] is not typically considered invertible for sin(x) without further restrictions, as the function will not pass the horizontal line test within this interval. Here, the sine function commences at zero, rises to a maximum, falls to a minimum, and rises back to zero, meaning certain y-values will match with two different x-values within this domain.