Domain and range of the function (arctan(ln(sqrtx)-1)))^3

[Emworthington]

Homework Statement

f(x) = (arctan(ln(sqrtx - 1)))^3

Homework Equations

domain of arctan: all real numbers
range of arctan: -∏/2, ∏/2

The Attempt at a Solution

I know that domain is x>0 when x ≠ 1, because I need a positive number to go under the radical and the natural log of 0 is undefined. For the range, then, I worked inwards through the parentheses and then set lnsqrt(x) -1 greater than -pi/2 and less than pi/2. Still, I think I may have made a mistake because my answers keep coming out different. Also, I don't know the effect that the cubed on the whole equation has. Any help to clarify would be greatly appreciated.

 

[SammyS]

Homework Statement

f(x) = (arctan(ln(sqrtx - 1)))^3

Homework Equations

domain of arctan: all real numbers
range of arctan: -∏/2, ∏/2

The Attempt at a Solution

I know that domain is x>0 when x ≠ 1, because I need a positive number to go under the radical and the natural log of 0 is undefined. For the range, then, I worked inwards through the parentheses and then set ln(sqrt(x) -1) greater than -pi/2 and less than pi/2. Still, I think I may have made a mistake because my answers keep coming out different. Also, I don't know the effect that the cubed on the whole equation has. Any help to clarify would be greatly appreciated.

It's best to work from the inside out.

In general, the Domain of f(g(x)) is: all values of x in the domain of g, such that g(x) is in the domain of f .

Finding the range can be a bit trickier.

Is your function $f(x)=\arctan(\ln(\sqrt{x}-1)\,)\,?$

Or is it $f(x)=\arctan(\ln(\sqrt{x-1}\,)\,)\,?$