Yahoo Answers: Answers and Comments for Domain and range of g(x) = sin1(2x + 3)...? [Mathematics]
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enUS
Sat, 12 Sep 2009 19:27:51 +0000
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Yahoo Answers: Answers and Comments for Domain and range of g(x) = sin1(2x + 3)...? [Mathematics]
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From Anonymous: The arcsine is the inverse of the sine meaning...
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Sat, 12 Sep 2009 19:53:45 +0000
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 codomain.
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]

From John F: arcsin [ u (x) ] is defined for  1 ≤ u (x) ≤...
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Sat, 12 Sep 2009 19:37:56 +0000
arcsin [ u (x) ] is defined for  1 ≤ u (x) ≤ 1
⇔
arcsin (2x + 3) is defined for :
 1 ≤ 2x + 3 ≤ 1
 1  3 ≤ 2x ≤ 1  3
 4 ≤ 2x ≤  2
 2 ≤ x ≤  1
⇔
Domain of arcsin (2x  3) = [  2 ;  1 ]
*******
arcsin ( 1) =  π/2
arcsin (1) = π/2
⇔
Range = [  π/2 ; π/2 ]
*******

From hopkin: Domain Of Sin Inverse
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Fri, 21 Oct 2016 11:55:06 +0000
Domain Of Sin Inverse