(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The forward-back function is

[tex]f (t) = 2t[/tex] for [tex]0\leq{t}\leq{3}[/tex] ,

[tex]f(t)= 12-2t[/tex] for [tex]3\leq{t}\leq{6}[/tex]. Graph [tex]f(f(t))[/tex] and find

itsfour-partformula. First try t = 1.5 and 3.

3. The attempt at a solution

There are four possible composite functions from the two given functions:

[tex]2(2(t))=4t[/tex]

[tex]12-2(2(t))=-4t+12[/tex]

[tex]2(12-2t)=-4t+24[/tex]

[tex]12-2(12-2t)=4t-12[/tex]

Since the domain of [tex]f (t) = 2t[/tex] is from 0 to 3 and the domain of [tex]f(t)= 12-2t[/tex] 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:

[tex]2(2(t))[/tex] for [tex]0\leq{t}\leq{3}[/tex]

[tex]12-2(12-2t)[/tex] for [tex]3\leq{t}\leq{6}[/tex]

The answer in the back of the book gives:

[tex]2(2t)[/tex] for [tex]0\leq{t}\leq{1.5}[/tex]

[tex]12-4t[/tex] for [tex]1.5\leq{t}\leq{3}[/tex]

[tex]12-2(12-2t)[/tex] for [tex]3\leq{t}\leq{4.5}[/tex]

[tex]2(12-2t)[/tex] for [tex]4.5\leq{t}\leq{6}[/tex]

I also don't see the relevance of "First try t = 1.5 and 3".

Any help will be appreciated.

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# Homework Help: Composite functions and domains

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