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Thread: Functions of functions

  1. #1
    Jul 2008

    Functions of functions

    its easy to plug a function into a function, we did that in algebra.
    f(x)=x^2 g(x) = (x+2) fog(x)= x^2+4x+4

    but how about going the other way
    if fog(x)= 3x^2+2x+8 and g(x) = (x^2+4x), what is f(x)?

    I never learned this in precalc or algebra, which branch or course deals with this topic? What is the process for solving these kind of problems?
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  2. #2
    Newbie Catherine Morland's Avatar
    Jul 2008
    Try expressing f\circ g(x) in terms of g(x).

    g(x)=x^2+4x\implies g(x)=(x+2)^2-4\implies x=-2+\sqrt{g(x)+4} or x=-2-\sqrt{g(x)+4}

    \therefore f(g(x))=3(-2+\sqrt{g(x)+4})^2+2(-2+\sqrt{g(x)+4})+8 or 3(-2-\sqrt{g(x)+4})^2+2(-2-\sqrt{g(x)+4})+8

    \therefore f(x)=3(-2+\sqrt{x+4})^2+2(-2+\sqrt{x+4})+8 or 3(-2-\sqrt{x+4})^2+2(-2-\sqrt{x+4})+8
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  3. #3
    Senior Member JaneBennet's Avatar
    Dec 2007
    Be careful here! You get two different functions f(x) depending on what domain you choose for g(x).

    g(x)+4=(x+2)^2 \Rightarrow \sqrt{g(x)+4}=\begin{cases}x+2 & x\ge-2\\<br />
-x-2 & x\le-2\end{cases}\ \Rightarrow\ x=\begin{cases}-2+\sqrt{g(x)+4} & x\ge-2\\<br />
-2-\sqrt{g(x)+4} & x\le-2\end{cases}

    If you choose [-2,\infty) as your domain for g(x), then f((g(x))=3\left(-2+\sqrt{g(x)+4}\right)^2+2\left(-2+\sqrt{g(x)+4}\right)+8 and so f(x)=3\left(-2+\sqrt{x+4}\right)^2+2\left(-2+\sqrt{x+4}\right)+8.

    If you choose (-\infty,-2] as your domain for g(x), then f((g(x))=3\left(-2-\sqrt{g(x)+4}\right)^2+2\left(-2-\sqrt{g(x)+4}\right)+8 \Rightarrow f(x)=3\left(-2-\sqrt{x+4}\right)^2+2\left(-2-\sqrt{x+4}\right)+8.

    (In both cases, the domain of f(x) is [-4,\infty).)

    However, if you choose \mathbb{R} as your domain for g(x), then, unfortunately, it will be impossible to determine f(x) as a function. ( f(x) would be a one-to-many relation and therefore not a function.)
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