The arcsine is the inverse of the sine meaning that the values of sine on the x - axis become the values of arcsine on the y - axis and the values of sine on the y - axis become the values of arcsine on the x - axis.

Since sine has a range of -1 <= y <= 1, then the domain of arcsine will be -1 <= x <= 1. However, this isn't arcsin(x), but arcsin(2x + 3).

{<= is the "less than or equal to" symbol}

So the domain is the set of all possible x values that can be substituted into the function. Therefore, since arcsine has a maximum of 1 and a minimum of -1 in the domain, arcsin(2x + 3) has a domain of

-1 <= 2x + 3 <= 1

Solving the inequality by subtracting -3 on both sides

-1 - 3 <= 2x <= 1 - 3

-4 <= 2x <= -2

Divide both sides by 2

-2 <= x <= -1

So the domain of arcsin(2x + 3) is [-2, -1] or

D = {x | -2 <= x <= -1}

Now for the range, arcsine has multiple y values for each x, for example

arcsin(0) = 2π = 0 = π

Which means that arcsine must be restricted to it's principal branch, that is where each x value for arcsine corresponds to only one y value. We have to do this, otherwise arcsine would break the definition of a function. That is for each x value in the function g(x)'s domain, must map to AT MOST ONE y value in the co-domain.

The principal branch of arcsine is -π/2 <= y <= π/2, but this is ONLY for arcsin(x), we want arcsin(2x + 3).

Since the domain is restricted too -2 <= x <= -1, then the range is the set of all values g(x) takes when x takes values of the domain. So, when you plug in the maximum value (x = -1) into the equation, you get

x = -1

arcsin(-2 + 3) = arcsin(1) = π/2

And the minimum value in the domain (x = -2)

x = -2

arcsin(-4 + 3) = arcsin(-1) = -π/2

So the range is [-π/2, π/2]

FINAL ANSWER:

Range = [-π/2, π/2]

Domain = [-2, -1]

Source(s):
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

http://www.analyzemath.com/DomainRange/DomainRange.html

Anonymous
· 10 years ago