How to find the domain of arccos(square root of x)

[Dark Horse]

Hey guys, ho do you find the domain of inverse trig functions that have functions contained within them? eg. arccos(square root of x) etc?

[qshyrn]

its a composition function dom = dom inner intersection dom outer   = [0,infinity) intersection [-1,1]      so u get [0,1]

[theuncle]

not necessarily qshryn...
for instance the domain of
is []
which is not the intersection of the domains of and arccos(x)

A better way to look at it is that the domain f(g(x)) is the domain of g(x) that gives a range of g(x) within the domain of f(x).
In other words you've gotta make sure that the values that g(x) outputs are within the domain of f(x), and restrict x accordingly

[qshyrn]

not necessarily qshryn...
for instance the domain of
is []
which is not the intersection of the domains of and arccos(x)

A better way to look at it is that the domain f(g(x)) is the domain of g(x) that gives a range of g(x) within the domain of f(x).
In other words you've gotta make sure that the values that g(x) outputs are within the domain of f(x), and restrict x accordingly

yeah, true im mistaken.    
       ran of sqroot(x) must be in [-1,1]      -1<=sqrroot(x)<=1    by definition sqrroot(x)is already greater or equal to 0,    so it becomes    0<=sqrroot(x)<=1   then u square all the terms getting     0<=(x)<=1

i think thats right?

[theuncle]

yeah that's it