http://mathworld.wolfram.com/Domain.html
first off, the above, second off, smack your math teacher around and tell the lousy moron to use english and stop trying to impress his students with his knowledge of learned jargon. He really isn't proving anything except that he's a lousy human being to a bunch of kids.
Now, using the most common description (I never use this archaic notation, so really I just decided to look this stuff up on the fly)That of the set of values over which our functiosn are defined. The composition of g^o h is g(f(x)) via,
http://mathworld.wolfram.com/Composition.html
And thus, the domain of g(f(x)) is the set of values over which g(f(x)) is defined? Make sense? I doubt it, but let's continue,
EQN 1; g(x) = |x| - 4
EQN 2; h(x) = sqrt(-2 - x)
EQN 1 -> EQN 2 -> EQN3; g(h(x))=|h(x)|-4
Or, plugging in the direct equation
EQN 3; g(h(x))=|
sqrt(-2 - x)|-4
Which becomes,
EQN 3 -> g(h(x))=+sqrt(-2-x)-4
As we are only conserned with the real values, the sqare root cannot have any negative values inside, and as a result,
EQN 4; -2-x=>0
Or,
EQN 4;
x<=-2
This last part being the solution, or,
x is less then or equal to negative two in order for g(h(x)) to be real.
