Let #F(x)=6-x# and #g(y)=sqrty#, how do you find each of the compositions and domain and range?

Question

Chloe Alexander · Accepted Answer

undefined To find the composition of functions F(g) and g(F), first substitute g(y) into F(x) and then substitute F(x) into g(y).

1. F(g):
F(g(y)) = 6 - √y

2. g(F):
g(F(x)) = √(6 - x)

Now, let's determine the domain and range for each composition:

For F(g):
Domain: The domain of F(g) depends on the domain of g(y), which is all non-negative real numbers.
Range: The range of F(g) depends on the range of F(x), which is all real numbers.

For g(F):
Domain: The domain of g(F) depends on the domain of F(x), which is all real numbers.
Range: The range of g(F) depends on the range of g(y), which is all non-negative real numbers.

Madelyn Anderson · Answer

#F# is a polynomial and therefore, it's domain is all real numbers. Since it is an odd polynomial, it's range is also all real numbers.  #g# has a non-integer in the exponent, so a negative input doesn't make any sense, hence it's domain is all positive real numbers. We also know that squareroots are always positive, so it's range is also all real numbers.  There are two ways to compose these two: 
#F(g(x)) = 6 - sqrt(x) #
This function has the same domain as #g#, but can only output numbers lower than 6, hence its range is #[6, -infty)#.  #g(F(x)) = sqrt(6-x)#
We must make sure that #6-x# is a positive value, so #x <=6#. Similarly, the squareroot is always non-negative and we can get 0, so its range is positive reals.  In summary: 
#F: (-infty, infty) -> (-infty, infty)#
#g: [0, infty) -> [0, infty) #
#F(g): [0, infty) -> (-infty, 6] #
#g(F): (-infty, 6] -> [0, infty) #