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Math Help - Square Root Function

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    Square Root Function

    Create a square root function with:

    (a) domain of [3, \infty) and a range of (-\infty, 5].

    Would this be right? f(x)=-\sqrt{x+3}+5

    And how would you create a square root function that has a domain of all real numbers?
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by chrozer View Post
    Create a square root function with:

    (a) domain of [3, \infty) and a range of (-\infty, 5].

    Would this be right? f(x)=-\sqrt{x+3}+5
    It would be better if it was f(x)=-\sqrt{x{\color{red}-}3}+5


    And how would you create a square root function that has a domain of all real numbers?
    \sqrt{x^2+\dots}

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    Quote Originally Posted by Moo View Post
    Hello,



    It would be better if it was f(x)=-\sqrt{x{\color{red}-}3}+5




    \sqrt{x^2+\dots}

    Duh....that was a stupid mistake. And thanks for the second part.

    Two more things...how would you determine the domain and range of f(x)=\sqrt{ax+b}+c in terms of a, b, and c.

    And would f^{-1}(x)=\frac {(x-c)^2-b}{a} be the inverse of f(x)=\sqrt{ax+b}+c?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chrozer View Post
    Duh....that was a stupid mistake. And thanks for the second part.

    Two more things...how would you determine the domain and range of f(x)=\sqrt{ax+b}+c in terms of a, b, and c.

    And would f^{-1}(x)=\frac {(x-c)^2-b}{a} be the inverse of f(x)=\sqrt{ax+b}+c?
    For \sqrt{ax+b}+c to be \in\mathbb{R}

    ax+b\geq{0}\Rightarrow{x\geq\frac{-b}{a}}

    and for the second one since f(x)=\sqrt{ax+b}+c\Rightarrow{x=\sqrt{af^{-1}(x)+b}+c}

    Now just solve for f^{-1}(x)
    Last edited by Mathstud28; May 17th 2008 at 11:33 AM.
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    Quote Originally Posted by Mathstud28 View Post
    For \sqrt{ax+b}+c to be \in\mathbb{R}

    ax+b>0\Rightarrow{x>\frac{-b}{a}}

    and for the second one since f(x)=\sqrt{ax+b}+c\Rightarrow{x=\sqrt{af^{-1}(x)+b}+c}

    Now just solve for f^{-1}(x)
    Ahh I see. I got the second one right, how would you determine the range of the 1st question? And also can the domain equal zero?
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    Moo
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    Quote Originally Posted by chrozer View Post
    Ahh I see. I got the second one right, how would you determine the range of the 1st question? And also can the domain equal zero?
    For the range :

    You know that \sqrt{ax+b} goes from 0 to +\infty

    So \sqrt{ax+b}+c goes from 0+c=c to +\infty

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    Quote Originally Posted by Moo View Post
    For the range :

    You know that \sqrt{ax+b} goes from 0 to +\infty

    So \sqrt{ax+b}+c goes from 0+c=c to +\infty

    So all together the domain would be x \geq \frac{-b}{a} and the range would be y \geq c. Right?
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  8. #8
    Moo
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    Quote Originally Posted by chrozer View Post
    So all together the domain would be x \geq \frac{-b}{a} and the range would be c \leq y \leq +\infty. Right?
    Correct for the domain.
    For the range, it's like c \leq y
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  9. #9
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    Quote Originally Posted by Moo View Post
    Correct for the domain.
    For the range, it's like c \leq y
    Oh I see. Thnx alot.
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