Confused about domain and range

**zhupolongjoe** · September 21st 2010, 08:16 AM

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.

**Ulysses** · September 21st 2010, 08:33 AM

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

Thats how I see it.

**Plato** · September 21st 2010, 08:45 AM

Originally Posted by Ulysses

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

@Ulysses, did you notice that the function to cubed?

**Ulysses** · September 21st 2010, 08:46 AM

Oh, no, I didn't, sorry.

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

**Plato** · September 21st 2010, 09:00 AM

Originally Posted by Ulysses

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

That is correct. So what is the range?

**zhupolongjoe** · September 21st 2010, 09:48 AM

So would the range be (-infinity, pi/2)?

**Plato** · September 21st 2010, 10:08 AM

Consider the fact that $\arctan$ is a bounded function:
$\displaystyle<br /> \left( {\left( { - \frac{\pi }<br /> {2}} \right)^3 ,\left( {\frac{\pi }<br /> {2}} \right)^3 } \right)$

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