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Thread: Transformations

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    23

    Transformations

    Given the parent functions;
    $\displaystyle
    f(x)=c
    $
    $\displaystyle
    f(x)=x
    $
    $\displaystyle
    f(x)=x^2
    $
    $\displaystyle
    f(x)=x^3
    $
    $\displaystyle
    f(x)=x^n
    $
    $\displaystyle
    f(x)=|x|
    $
    $\displaystyle
    f(x)=b^x
    $
    $\displaystyle
    f(x)=\sqrt{x}
    $
    $\displaystyle
    f(x)=1/x
    $
    $\displaystyle
    f(x)=[x]
    $

    I'm asked to rewrite these functions in terms of thier parent function. I've shown what I've done, and marked ones that I'm pretty sure I messed up on. If anyone could buzz over these and still be willing to hand out a couple pointers, that would be awesome.

    $\displaystyle
    g(x)=3|x-2| => g(x)= 3f(x-2), ||||where f(x)=x
    $
    $\displaystyle
    g(x)=-x^2+4 => g(x)= -f(x)+4, ||||where f(x)=x^2
    $
    $\displaystyle
    *g(x)=(x-4)^3 => g(x)= f((x-4)^3), ||||where f(x)=x^3
    $
    $\displaystyle
    *g(x)=3^{-x}-2 => g(x)= 3f(-x)-2, ||||where f(x)=b^x
    $
    $\displaystyle
    *g(x)=3/x+2 => g(x)=3/f(x+2), ||||where f(x)=1/x
    $
    $\displaystyle
    g(x)=-[2x] => g(x)=-f(2x), ||||where f(x)=[x]
    $
    $\displaystyle
    g(x)=-4\sqrt{x-2} => g(x)=-4f(x-2), ||||where f(x)=\sqrt{x}$


    *I'm not sure If I've done these ones right.
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  2. #2
    Super Member

    Joined
    May 2006
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    Lexington, MA (USA)
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    Hello, EyesForEars!

    Given the parent functions:

    . . $\displaystyle \begin{array}{ccccc}f(x)\,=\,c &f(x),=\,x & f(x)\,=\,x^2 & f(x)\,=\,x^3 & f(x)\,=\,x^n \\ \\[-3mm]
    f(x)\,=\,|x| & f(x)\,=\,b^x & f(x)\,=\,\sqrt{x} & f(x)\,=\,\frac{1}{x} & f(x)\,=\,[x] \end{array}$

    I'm asked to rewrite these functions in terms of thier parent function.
    I've shown what I've done, and marked ones that I'm pretty sure I messed up on.

    $\displaystyle g(x)\,=\,3|x-2|\quad\Rightarrow\quad g(x)\,=\, 3f(x-2)\;\text{ where }f(x)\,=\,x$

    $\displaystyle g(x)\,=\,-x^2+4\quad\Rightarrow\quad g(x)\,=\, -f(x)+4\;\text{ where }f(x)\,=\,x^2$

    $\displaystyle g(x)\,=\,(x-4)^3 \quad\Rightarrow\quad g(x)\,=\, f((x-4)^3)\;\text{ where }f(x)\,=\,x^3$ .. . . no

    $\displaystyle g(x)\,=\,3^{-x}-2 \quad\Rightarrow\quad g(x)\,=\, 3f(-x)-2\;\text{ where }f(x)\,=\,b^x$ . . . . Right!

    $\displaystyle g(x)\,=\,\frac{3}{x+2} \quad\Rightarrow\quad g(x)\,=\,\frac{3}{f(x+2)}\;\text{ where }f(x)\,=\,1/x$ . . . . no

    $\displaystyle g(x)\,=\,-[2x] \quad\Rightarrow\quad g(x)\,=\,-f(2x)\;\text{ where }f(x)\,=\,[x]$

    $\displaystyle g(x)\,=\,-4\sqrt{x-2} \quad\Rightarrow\quad g(x)\,=\,-4f(x-2)\;\text{ where }f(x)\,=\,\sqrt{x}$

    If $\displaystyle f(x) \:=\:x^3$, then $\displaystyle f(x-4) \:=\:(x-4)^3$

    . . Therefore: .$\displaystyle g(x) \:=\:(x-4)^3 \:=\:f(x-4)$


    If $\displaystyle f(x) \:=\:\frac{1}{x}$, then $\displaystyle f(x+2) \:=\:\frac{1}{x+2}$
    . . Therefore: .$\displaystyle g(x) \:=\:\frac{3}{x+2} \:=\:3f(x+2)$

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  3. #3
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    Awesome.
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