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Math Help - 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?
    f(x)=\sqrt{\frac{x-1}{x+1}} <br />

    Thanks.
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  2. #2
    MHF Contributor
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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?
    f(x)=\sqrt{\frac{x-1}{x+1}} <br />

    Thanks.
    Hi

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

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

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

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

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

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

    The domain is )-\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 : [0 ; 1( \cup )1 ; +\infty(
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