Composite functions and domains

1. Feb 26, 2010

bacon

1. The problem statement, all variables and given/known data

The forward-back function is
$$f (t) = 2t$$ for $$0\leq{t}\leq{3}$$ ,
$$f(t)= 12-2t$$ for $$3\leq{t}\leq{6}$$. Graph $$f(f(t))$$ and find
its four-part formula. First try t = 1.5 and 3.

3. The attempt at a solution
There are four possible composite functions from the two given functions:

$$2(2(t))=4t$$

$$12-2(2(t))=-4t+12$$

$$2(12-2t)=-4t+24$$

$$12-2(12-2t)=4t-12$$

Since the domain of $$f (t) = 2t$$ is from 0 to 3 and the domain of $$f(t)= 12-2t$$ is from 3 to 6 it seems to me that only the first and fourth functions are valid. Functions two and three have a part of the composite function applied outside its defined domain. Instead of a four part formula I get a two part one:

$$2(2(t))$$ for $$0\leq{t}\leq{3}$$

$$12-2(12-2t)$$ for $$3\leq{t}\leq{6}$$

The answer in the back of the book gives:

$$2(2t)$$ for $$0\leq{t}\leq{1.5}$$

$$12-4t$$ for $$1.5\leq{t}\leq{3}$$

$$12-2(12-2t)$$ for $$3\leq{t}\leq{4.5}$$

$$2(12-2t)$$ for $$4.5\leq{t}\leq{6}$$

I also don't see the relevance of "First try t = 1.5 and 3".
Any help will be appreciated.

2. Feb 26, 2010

epenguin

The relevance is just so that you see whether inside its domain it's a continuous function. though maybe with sharp corners (if the slope of the function is not an everywhere continuous function). You have a formula f(t) = one thing for a stretch up to t=3 but another thing for $$t\leq{3}$$. Can you see intuitively that if the two formulae give the same result for t=3 the function is continuous, and if not there is a discontinuity?

And what is the relevance of that? Well in dealing with a function you'll almost always want to know what it looks like - you would be naturally led a the start to look at f(x) at the break point t=3. As for 1.5 - well you'll see when you have calculated f(f(t)).

Last edited: Feb 27, 2010