Find the domain of h(x) = [arcsin ( (2x-1)/(1-x) )]^2 - 2

[bookerdewitt12345]

How do i start this problem?

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[Deleted member 4993]

How do i start this problem?
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Please translate to English.

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[stapel]

How do i start this problem?

. . . . .Odredite domenu funkcije:

. . . . . . .\(\displaystyle h(x)\, =\, \left(\, \arcsin\, \left(\dfrac{2x\, -\, 1}{1\, -\, x}\right)\, \right)^2\, -\,2\)

When I typed the text into "Google Translate", it returned the text "specify the domain functions", and indicated that the language "detected" was "Croatian". Is the following correct?

. . . . .Find the domain of the function:

. . . . . . .\(\displaystyle h(x)\, =\, \left(\, \arcsin\, \left(\dfrac{2x\, -\, 1}{1\, -\, x}\right)\, \right)^2\, -\,2\)

If so, then one would start by checking the domain of the polynomial fraction, because one is not allowed to divide by zero. Then one would turn to the domain of the inverse sine function.