What is the domain of #(f@g)(x)#?

Question

Benjamin Achtenberg · Accepted Answer

If #g:A->B# and #f:B->C#, then the domain of #f@g# is

#bar(g)^(-1)@bar(f)^(-1)(C)#

using the notation described below...

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 pre-image 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 #f@g# is simply #bar(g)^(-1)(bar(f)^(-1)(C)) = (bar(g)^(-1)@bar(f)^(-1))(C)#

Sadie Ackerman · Answer

undefined To determine the domain of $ (f@g)(x) $, where $ f $ and $ g $ are functions, we need to consider the domain restrictions imposed by both functions individually and any restrictions imposed by the operation $ g \circ f $.

Firstly, we need to ensure that $ x $ belongs to the domain of $ f $, such that $ f(x) $ is defined. Then, we need to ensure that the output of $ f(x) $ belongs to the domain of $ g $, so that $ g(f(x)) $ is well-defined.

In simpler terms, for each function $ f $ and $ g $, we need to identify any values of $ x $ that would make $ f(x) $ undefined or produce outputs that are not in the domain of $ g $.

Once we've identified any such values, the domain of $ (f@g)(x) $ will be the set of all $ x $ values that satisfy these conditions.

In summary, to find the domain of $ (f@g)(x) $, we:
1. Identify the domain of $ f $ and $ g $.
2. Determine any values of $ x $ for which $ f(x) $ is undefined or produces outputs not in the domain of $ g $.
3. Exclude these values from the domain of $ (f@g)(x) $.

The resulting set of values will be the domain of $ (f@g)(x) $.