Domain of a function problem: inverse sin(sqrt(2x/1-2x)

[mannistrang]

Hello, I have a quick question regarding the domain of a function. In the function inverse sin(sqrt(2x/1-2x) the domain should be [0,1/2] since the numbers cant be negative and anything greater than 1/2 would be negative. My question is that when I graph it, the domain only goes up to 1/4. Can anyone explain to me why this is?

Thanks in advance

 

[Deleted member 4993]

Hello, I have a quick question regarding the domain of a function. In the function inverse sin(sqrt(2x/1-2x) the domain should be [0,1/2] since the numbers cant be negative and anything greater than 1/2 would be negative. My question is that when I graph it, the domain only goes up to 1/4. Can anyone explain to me why this is?

Thanks in advance

If you put x = 0.3, what do you get as the argument of arcsin (for this problem)?

 

[Otis]

inverse sin(sqrt(2x/[1-2x])

the domain should be [0,1/2) since the numbers cant be negative

Note the change in red above; x cannot equal 1/2 because that leads to division by zero.

The inverse sine function is also called arcsin.

arcsin(sqrt(2x/[1-2x]) is a composite function: the sqrt() function can be considered the inside function, and the outer function is the arcsin() function.

[0,1/2) is not the domain of the composite function. It's just a restricted domain on the inner function, to assure that the inner function outputs non-negative values, which in turn are input into the outer function.

Think about the domain of arcsin(x). You need to adjust the domain of your inner function so that the inner function outputs values that are in the domain of arcsin(x).

What is the domain of arcsin(x)?