Domains of composition functions: I am having a hard time with this! An example of one problem in my book: "Find f o g ( f(g(x)) ) g o f, f o f, g o g for the following problem(s): f(x) = x / 1+x [x divided by 1+x ] g(x) = sin 2x for fog I got sin2x / 1 + sin2x (sin2x divided by 1+sin2X) which means that sin 2x CANNOT equal -1. When you solve for x, you get 7pi/6 since the sin of 7pi/6 is -1/2 (negative one half). Is the domain simply All Real Numbers except -1/2? Or
All real Numbers except -1/2 and -1? I am confused. For gof I got Sin2 (x/ 1+x) Which seems to say the domain is all
Real numbers EXCEPT for -1. Is this the entire answer for the domain? For fof I got x/ 1+2x (x divided by 1+2x). 2x cant equal -1 and x cant equal -1/2.... what is the domain?? For gog I got sin2 (sin2x). The domain would be all Real
Numbers (right?).
Finding the domain of a composite function consists of two steps:
(a) Find the domain of the "inside" (input) function. If there are any restrictions on the domain, keep them.
(b) Construct the composite function. Find the domain of this new function. If there are restrictions on this domain, add them to the restrictions from (a). If there is an overlap, use the more restrictive domain.
(1) fog(x) = f(g(x)) = f(sin 2x) = sin 2x / (1 + sin 2x)
Domain of g(x) is all real numbers
Domain of fog(x) is sin 2x ≠ -1, that is 2x ≠ arcsin(-1), that is 2x ≠ n π + (-1)^n (-π /2), that is x ≠ {n π/2 + (-1)^n (-π /4)}
Answer: Domain is {x | x ≠ {n π/2 + (-1)^n (-π /4)}, that is, {x | x ≠ -π/4, x ≠ 3 π/4, x ≠ 7 π/4, …}
(2) gof(x) = g(f(x)) = g(x/(1 + x)) = sin(2x/(1 + x))
Domain of f(x) is {x | x ≠ -1}
Domain of gof(x) is all real numbers
Answer: Domain is {x | x ≠ -1}
(3) fof(x) = f(f(x)) = f(x/(1 + x)) = x/(1 + 2x)
Domain of f(x) is {x | x ≠ -1}
Domain of fof(x) is {x | x ≠ -1/2}
Answer: Domain is {x | x ≠ -1/2 and x ≠ -1}
(4) gog(x) = g(g(x)) = g(sin 2x) = sin(2(sin 2x))
Domain of g(x) is all real numbers
Domain of gog(x) is all real numbers
Answer: Domain is all real numbers.