well, that's just multiplying the "2" in but, yes, it is a little "simplified".
Then I believe the domain would be {x|x not= 0}.[/QUOTE]
No. You cannot divide by 0 but here, the denominator is 1+ x. That is 0 when x= -1, not when x=0. If x= 0, that would be sin(0)= 0 which is well defined.
No. This is the product, fg(x). The composition, as dwsmith says.
The domain is "sin(2x) not = -1" what values of x satisfy that?Domain {x|x not= -1)
Multipying both numerator and denominator by 1+ x makes that just as you have!
Very good! Although the formula you give is "calculable" at x= -1, f is not and you must be able to do that for f(f(-1))!Domain {x|x not=-1,-2}
Yes, that is correct. Your only problem was fof and since you got the other correct, I assume that was just a careless errror.
Domain = All real numbers