Function Composition
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Key Questions

To compose a function is to input one function into the other to form a different function. Here's a few examples.
Example 1: If
#f(x) = 2x + 5# and#g(x) = 4x  1# , determine#f(g(x))# This would mean inputting
#g(x)# for#x# inside#f(x)# .#f(g(x)) = 2(4x 1) + 5 = 8x 2 + 5 = 8x + 3# Example 2: If
#f(x) = 3x^2 + 12 + 12x# and#g(x) =sqrt(3x)# , determine#g(f(x))# and state the domainPut
#f(x)# into#g(x)# .#g(f(x)) = sqrt(3(3x^2 + 12x + 12))# #g(f(x)) = sqrt(9x^2 + 36x + 36)# #g(f(x)) = sqrt((3x + 6)^2)# #g(f(x)) = 3x + 6# The domain of
#f(x)# is#x in RR# . The domain of#g(x)# is#x > 0# . Hence, the domain of#g(f(x))# is#x > 0# .Example 3: if
#h(x) = log_2 (3x^2 + 5)# and#m(x) = sqrt(x + 1)# , find the value of#h(m(0))# ?Find the composition, and then evaluate at the given point.
#h(m(x)) = log_2 (3(sqrt(x + 1))^2 + 5)# #h(m(x)) = log_2 (3(x + 1) + 5)# #h(m(x)) = log_2 (3x + 3 + 5)# #h(m(x)) = log_2 (3x + 8)# #h(m(2)) = log_2 (3(0) + 8)# #h(m(2)) = log_2 8# #h(m(2)) = 3# Practice exercises
For the following exercises:
#f(x) = 2x + 7, g(x) = 2^(x  7) and h(x) = 2x^3  4# a) Determine
#f(g(x))# b) Determine
#h(f(x))# c) Determine
#g(h(2))# Hopefully this helps, and good luck!

It is, if the following works:
#([email protected]([email protected]))(x) = (([email protected])@h)(x)# That is, if:
#f(x) = "something"#
#g(h(x)) = ([email protected])(x) = "something else"#
#f(g(x)) = ([email protected])(x) = "something else again"#
#h(x) = "something else yet again"# ...and you can use these together to satisfy the first expression, then they are associative. Let:
#f(x) = 2x#
#g(x) = x^2#
#h(x) = x^3# Thus:
#g(h(x)) = g(x^3) = (x^3)^2 = x^6# #f(g(x)) = g(x^2) = 2(x^2) = 2x^2# Then:
#([email protected]([email protected]))(x) = f(x^6) = 2(x^6) = color(blue)(2x^6)# #(([email protected])@h)(x) = f(g(x^3)) = f((x^3)^2) = f(x^6) = color(blue)(2x^6)# Therefore they are associative.

Answer:
If
#g:A>B# and#f:B>C# , then the domain of#[email protected]# is#bar(g)^(1)@bar(f)^(1)(C)# using the notation described below...
Explanation:
If
#g# is a function that maps some elements of a set#A# to elements of a set#B# , then the domain of#g# is the subset of#A# for which#g(a)# is defined.More formally:
#g sube A xx B :# #AA a in A AA b_1, b_2 in B# #((a, b_1) in g ^^ (a, b_2) in g) => b_1 = b_2# Use the notation
#2^A# to represent the set of subsets of#A# and#2^B# the set of subsets of#B# .Then we can define the preimage function:
#bar(g)^(1): 2^B > 2^A# by#bar(g)^(1)(B_1) = {a in A : g(a) in B_1}# Then the domain of
#g# is simply#bar(g)^(1)(B)# If
#f# is a function that maps some elements of set#B# to elements of a set#C# , then:#bar(f)^(1): 2^C > 2^B# is defined by#bar(f)^(1)(C_1) = {b in B : f(b) in C_1}# Using this notation, the domain of
#[email protected]# is simply#bar(g)^(1)(bar(f)^(1)(C)) = (bar(g)^(1)@bar(f)^(1))(C)#
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