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Thread: Help for domain and range

  1. #1
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    Help for domain and range

    Hi all,

    I am having trouble with the process for this problem.

    What is the domain and range for this function?
    $\displaystyle f(x)=\sqrt{\frac{x-1}{x+1}}
    $

    Thanks.
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  2. #2
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    Quote Originally Posted by romjke View Post
    Hi all,

    I am having trouble with the process for this problem.

    What is the domain and range for this function?
    $\displaystyle f(x)=\sqrt{\frac{x-1}{x+1}}
    $

    Thanks.
    Hi

    The domain is the set of x for which $\displaystyle f(x)=\sqrt{\frac{x-1}{x+1}}$ exists

    f(x) exists when $\displaystyle \frac{x-1}{x+1} \geq 0$

    For $\displaystyle x < -1$
    $\displaystyle x-1 < 0$ and $\displaystyle x+1 < 0$ therefore $\displaystyle \frac{x-1}{x+1} > 0$

    For x=-1 x+1 =0 therefore f(x) is not defined

    For $\displaystyle -1 < x < 1$
    $\displaystyle x-1 < 0$ and $\displaystyle x+1 > 0$ therefore $\displaystyle \frac{x-1}{x+1} < 0$

    For $\displaystyle x \geq 1$
    $\displaystyle x-1 \geq 0$ and $\displaystyle x+1 > 0$ therefore $\displaystyle \frac{x-1}{x+1} \geq 0$

    The domain is $\displaystyle )-\infty ; -1( \cup [1 ; +\infty($
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  3. #3
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    Thank you, but what about the range?
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  4. #4
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    Quote Originally Posted by romjke View Post
    Thank you, but what about the range?
    Calculate the derivative
    Find the variations of f
    Find the limits at the boundaries of the domain
    You will find the range : $\displaystyle [0 ; 1( \cup )1 ; +\infty($
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