Composite functions and domains

[bacon]

Homework Statement

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.

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.

 

[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)).