# Let F(x) = arcsin((x-2)/2) 1) what is the domain of F(x)? 2) What is the domain of F'(x)= 1/(sqrt(-(x-4)(x))) ?

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### 1 Answer

1) You need to remember that the domain of the arcsine function is the interval `[-1,1], ` hence, you need to solve the next inequality such that:

`-1 =< (x-2)/2 =< 1 => -1 =< x - 2 =<2 => -1 + 2 =< x - 2 + 2 =< 2 + 2`

`1 =< x =< 4`

**Hence, evaluating the domain of the given function F(x) yields `x in [1 ; 4].` **

2) Notice that the derivative of the function F(x) is `F'(x) = 1/(sqrt(1 - (x-2)^2/4))*((x-2)/2)'` such that:

`F'(x) = 1/(2sqrt(1 - (x-2)^2/4)) =>F'(x) = 1/(sqrt(4 - (x-2)^2))`

`F'(x) = 1/(sqrt((2-x+2)(2+x-2))) => F'(x) = 1/(sqrt(x(4-x)))`

Notice that the function F'(x) exists for all values of x that do not cancel the denominator. Moreover, the radicand needs to be larger than zero, hence ,you may find the domain of derivative solving the inequality `x(4-x)>0` such that:

`x(4-x)>0 => {(x > 0),(4-x>0):} => {(x > 0),(-x>-4):} => {(x > 0),(x<4):} => x in (0 , 4)`

**Hence, evaluating the domain of derivative of F(x) yields `x in (0 , 4).` **