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Math Help - Confused about domain and range

  1. #1
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    Confused about domain and range

    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.
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Ulysses View Post
    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?
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  4. #4
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    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.
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  5. #5
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    Quote Originally Posted by Ulysses View Post
    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?
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  6. #6
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    So would the range be (-infinity, pi/2)?
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  7. #7
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    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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