2
$\begingroup$

I am trying to determine the function $f \circ g$ where

$$ f(x) = \begin{cases} x+2, & \text{if }x\ < -1 \\ -x, & \text{if }-1 \leqslant x\leqslant 1 \\ x-2,& \text{if }x > 1 \end{cases} $$

and

$$ g(x) = \begin{cases} x-2, & \text{if }x\ < -1 \\ -x, & \text{if }-1 \leqslant x\leqslant 1 \\ x+2,& \text{if }x > 1 \end{cases} $$

I understand function composition. The problem I am having here is that I don't know how to determine what happens to the input of $f$ from the output of $g$. I am not looking for the answer, I just need understand exactly how a composition of piecewise functions works. I haven not been able to obtain anything useful from other sources. I really don't know where to start solving this exercise. Is there anybody that has any valuable input they can give me?

$\endgroup$
2
$\begingroup$

You can do this with cases.

For example. Let's say we want $f(g(x))$ for $x<-1$, then we have $f(g(x)) = f(x-2)$ since $g(x) = x-2$ for $x<-1$. Now since $x<-1$, this means that $x-2<-3$ which means that $f(x-2) = (x-2)+2 = x$. Now proceed with the other cases.

$\endgroup$
  • $\begingroup$ Thank you! What about the case $f(-x)$? Since $-1 \leqslant x \leqslant 1$, does this mean that $1 \geqslant -x \geqslant -1$ ? $\endgroup$ – Doug Ramsey Mar 1 '13 at 3:30
  • $\begingroup$ Yeah you have the right idea. $\endgroup$ – joshphysics Mar 1 '13 at 3:55
1
$\begingroup$

First split up the inputs to the composite function according the the cases for $g$. Within each case, look at the outputs: are they all in one of $f$’s cases, or not? If they are, life is simple: $f\circ g$ will have the same cases as $g$. If not, you’ll have to subdivide $g$’s cases.

Here, for instance, consider the case $x<-1$: $g(x)=x-2$, so $g(x)<-1$ for all $x<-1$, and calculating $(f\circ g)(x)$ will require using the first case of $f$. Something similar happens in the other cases, so with this pair of functions you won’t actually have to do and subdividing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.