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Math Help - Transformations

  1. #1
    Newbie
    Joined
    Oct 2008
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    23

    Transformations

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

    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.

    <br />
g(x)=3|x-2| => g(x)= 3f(x-2), ||||where f(x)=x<br />
    <br />
g(x)=-x^2+4 => g(x)= -f(x)+4, ||||where f(x)=x^2<br />
    <br />
*g(x)=(x-4)^3 => g(x)= f((x-4)^3), ||||where f(x)=x^3<br />
    <br />
*g(x)=3^{-x}-2 => g(x)= 3f(-x)-2, ||||where f(x)=b^x<br />
    <br />
*g(x)=3/x+2 => g(x)=3/f(x+2), ||||where f(x)=1/x<br />
    <br />
g(x)=-[2x] => g(x)=-f(2x), ||||where f(x)=[x]<br />
    <br />
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:

    . . \begin{array}{ccccc}f(x)\,=\,c &f(x),=\,x & f(x)\,=\,x^2 & f(x)\,=\,x^3 & f(x)\,=\,x^n \\ \\[-3mm]<br />
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.

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

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

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

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

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

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

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

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

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


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

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  3. #3
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    Awesome.
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