# Confused about domain and range

• Sep 21st 2010, 08:16 AM
zhupolongjoe
We need to find the domain and range of (arctan(ln(sqrt(x)-1)))^3

I have domain and range of ln(sqrt(x)-1) as D: x>1 and R: reals. Is it correct? The arctan throws me off. I am not sure..I know for arctan(x), the domain is R and the range is (-pi/2,pi/2), but here I am not sure.
• Sep 21st 2010, 08:33 AM
Ulysses
I'm agree for the domain, but why you say that the range are the reals? you actually know that $\displaystyle \frac{-\pi}{2}<arctan<\frac{\pi}{2}$, so the range must be between those values, and actually as you got the ln, you know that the range is $\displaystyle 0<arctan<\frac{\pi}{2}$

Thats how I see it.
• Sep 21st 2010, 08:45 AM
Plato
Quote:

Originally Posted by Ulysses
I'm agree for the domain, but why you say that the range are the reals? you actually know that $\displaystyle \frac{-\pi}{2}<arctan<\frac{\pi}{2}$, so the range must be between those values, and actually as you got the ln, you know that the range is $\displaystyle 0<arctan<\frac{\pi}{2}$

@Ulysses, did you notice that the function to cubed?
• Sep 21st 2010, 08:46 AM
Ulysses
Oh, no, I didn't, sorry.

I actually think that my last deduction about the natural logarithm its wrong, cause it tends to $\displaystyle -\infty$ as the inside of it tends to zero.
• Sep 21st 2010, 09:00 AM
Plato
Quote:

Originally Posted by Ulysses
I actually think that my last deduction about the natural logarithm its wrong, cause it tends to $\displaystyle -\infty$ as the inside of it tends to zero.

That is correct. So what is the range?
• Sep 21st 2010, 09:48 AM
zhupolongjoe
So would the range be (-infinity, pi/2)?
• Sep 21st 2010, 10:08 AM
Plato
Consider the fact that $\displaystyle \arctan$ is a bounded function:
$\displaystyle \displaystyle \left( {\left( { - \frac{\pi } {2}} \right)^3 ,\left( {\frac{\pi } {2}} \right)^3 } \right)$