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The domain of a composite function

According to my class work book the domain of a composite function, fg(x) has domain of g(x).

While this holds true for some composite functions, I don't believe it holds true for all.

Any assistance would be greatly appreciated.
Original post by Jglover653
According to my class work book the domain of a composite function, fg(x) has domain of g(x).

While this holds true for some composite functions, I don't believe it holds true for all.

Any assistance would be greatly appreciated.

You are right. Consider the inclusion function f:{1}→Rf: \{1\} \to \mathbb{R} by x↦xx \mapsto x, and g:[0,1]→Rg: [0,1] \to \mathbb{R} by x↦x−2x \mapsto x-2. Then f∘gf \circ g is not defined anywhere.
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Jglover653
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Original post by Smaug123
You are right. Consider the inclusion function f:{1}→Rf: \{1\} \to \mathbb{R} by x↦xx \mapsto x, and g:[0,1]→Rg: [0,1] \to \mathbb{R} by x↦x−2x \mapsto x-2. Then f∘gf \circ g is not defined anywhere.


Thank you very much for your response.

That being said, would the domain of g(x) feature in the domain of fg(x).

As in, if the domain of g(x) excludes a value, will the domain of fg(x) also exclude said value, but not limited to this single exclusion.

Considering g(x)=1/x would result in the exclusion of 0.
Original post by Jglover653
That being said, would the domain of g(x) feature in the domain of fg(x).

This is wrong (as witnessed by the example above), but your rephrasing is correct:

As in, if the domain of g(x) excludes a value, will the domain of fg(x) also exclude said value, but not limited to this single exclusion.

Considering g(x)=1/x would result in the exclusion of 0.


It is true that the domain of fg is contained within the domain of g.

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