Results 1 to 3 of 3

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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie Catherine Morland's Avatar
    Jul 2008
    Try expressing $\displaystyle f\circ g(x)$ in terms of $\displaystyle g(x)$.

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

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

    $\displaystyle \therefore f(x)=3(-2+\sqrt{x+4})^2+2(-2+\sqrt{x+4})+8$ or $\displaystyle 3(-2-\sqrt{x+4})^2+2(-2-\sqrt{x+4})+8$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member JaneBennet's Avatar
    Dec 2007
    Be careful here! You get two different functions $\displaystyle f(x)$ depending on what domain you choose for $\displaystyle g(x)$.

    $\displaystyle g(x)+4=(x+2)^2$ $\displaystyle \Rightarrow$ $\displaystyle \sqrt{g(x)+4}=\begin{cases}x+2 & x\ge-2\\
    -x-2 & x\le-2\end{cases}\ \Rightarrow\ x=\begin{cases}-2+\sqrt{g(x)+4} & x\ge-2\\
    -2-\sqrt{g(x)+4} & x\le-2\end{cases}$

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

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

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

    However, if you choose $\displaystyle \mathbb{R}$ as your domain for $\displaystyle g(x)$, then, unfortunately, it will be impossible to determine $\displaystyle f(x)$ as a function. ($\displaystyle f(x)$ would be a one-to-many relation and therefore not a function.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Apr 15th 2010, 05:50 PM
  2. Replies: 3
    Last Post: Feb 23rd 2010, 04:54 PM
  3. Replies: 11
    Last Post: Nov 15th 2009, 11:22 AM
  4. Replies: 7
    Last Post: Aug 12th 2009, 04:41 PM
  5. Replies: 1
    Last Post: Apr 15th 2008, 09:00 AM

Search Tags

/mathhelpforum @mathhelpforum