Function Composition

[raybies]

In this context (in working with a function under the operation of composition) when we raise a function to a power like f^2, this means (f∘f)(x). In other words, we apply the composition twice.

Similarly, we would say (f∘g)^2(x)=((f∘g)∘(f∘g))(x) and continue this way for any power.

1. Show that function composition is not commutative. That is, find a suitable f(x) and g(x) such that (f∘g)(x)≠(g∘f)(x)

 

[Deleted member 4993]

In this context (in working with a function under the operation of composition) when we raise a function to a power like f^2, this means (f∘f)(x). In other words, we apply the composition twice.

Similarly, we would say (f∘g)^2(x)=((f∘g)∘(f∘g))(x) and continue this way for any power.

1. Show that function composition is not commutative. That is, find a suitable f(x) and g(x) such that (f∘g)(x)≠(g∘f)(x)

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:
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Please share your work/thoughts about this problem.

 

[Steven G]

No, (f∘g)^2(x) =[(f∘g)(x)]^2

 

[Dr.Peterson]

In this context (in working with a function under the operation of composition) when we raise a function to a power like f^2, this means (f∘f)(x). In other words, we apply the composition twice.

Similarly, we would say (f∘g)^2(x)=((f∘g)∘(f∘g))(x) and continue this way for any power.

1. Show that function composition is not commutative. That is, find a suitable f(x) and g(x) such that (f∘g)(x)≠(g∘f)(x)

Just try a simple pair of functions, and see whether (f∘g)(x) = (g∘f)(x). If it is, try a slightly more complicated pair.

No, (f∘g)^2(x) =[(f∘g)(x)]^2

No, they are defining the notation. As they say, "in this context". See here: https://en.wikipedia.org/wiki/Function_composition#Functional_powers

As I read it, everything in the question is a quote from the source; none is what the OP thinks. Note that the notation is irrelevant to question #1, and is presumably used in subsequent questions.

 

[Steven G]

I thought that it might be the definition from the OP's book but wasn't sure. It is terrible notation.