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Math Help - Hitting bit of a snag on domain and range of functions.

  1. #1
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    Hitting bit of a snag on domain and range of functions.

    Hi again all,

    I am asked to find domain and range of an f(x) and the range of its inverse f^-1(x):

    f(x) = sqrt(2x - 5)

    So I found the domain of this to be x >or= 5/2. I'm not too good at getting range so I decided to find the domain of the inverse which would then be the range of the original. My inverse turned out to be:

    f^-1(x) = (x^2 + 5)/ 2

    So then it seems like the domain to the above inverse function would be all real numbers, but then I had trouble verifying that the range of my original was all real numbers cause how could the range of f(x) = sqrt(2x - 5) be a negative number?
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  2. #2
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    Quote Originally Posted by DannyMath View Post
    So then it seems like the domain to the above inverse function would be all real numbers, but then I had trouble verifying that the range of my original was all real numbers cause how could the range of f(x) = sqrt(2x - 5) be a negative number?
    I think you are on the right track.

    Given your original function is f(x)=\pm\sqrt{2x-5} it will have y \in\mathbb{R}

    But this is not a one to one function so its inverse does not exist.

    Restricting the domain to force this function to be one to one you get.

    Firstly

    f(x)=+\sqrt{2x-5},~y\in \mathbb{R}^- \cup \{0\}

    and then

    f(x)=-\sqrt{2x-5},~y\in \mathbb{R}^+ \cup \{0\}

    The best way to see this is via a sketch.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    I think you are on the right track.

    Given your original function is f(x)=\pm\sqrt{2x-5} it will have y \in\mathbb{R}

    But this is not a one to one function so its inverse does not exist.

    Restricting the domain to force this function to be one to one you get.

    Firstly

    f(x)=+\sqrt{2x-5},~y\in \mathbb{R}^- \cup \{0\}

    and then

    f(x)=-\sqrt{2x-5},~y\in \mathbb{R}^+ \cup \{0\}

    The best way to see this is via a sketch.
    Ok well the original function I was given was indeed the:

    f(x)=+\sqrt{2x-5}

    function. But the problem I had was because I found that the inverse of this function had a domain of all real numbers, which made me conclude that the range of:

    f(x)=+\sqrt{2x-5}


    HAS to be all real numbers, but this doesn't seem to be the case, which is what confused me.
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