# Functions of functions

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• Aug 8th 2008, 08:42 AM
kd8bxz
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?
• Aug 8th 2008, 11:27 AM
Catherine Morland
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 $3(-2-\sqrt{x+4})^2+2(-2-\sqrt{x+4})+8$
• Aug 8th 2008, 12:44 PM
JaneBennet
Be careful here! You get two different functions $f(x)$ depending on what domain you choose for $g(x)$.

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

If you choose $[-2,\infty)$ 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.)