In this article I outline a systematic way of finding the domain of a composite function. A definition that can be used for this purpose follows:

\[D(f \circ g) = \{x|x\in D(g) \wedge g(x) \in D(f)\}\]

(Vaught, 1995:18)

Where \(D(\lambda) =\)  \(\text{the domain of }\lambda\)

The explanatory method which follows is to show how to use this definition in different examples.

Example 1

Solve \(D(\ln(\ln(\ln x)))\) .

Solution:

Let \(\ln(\ln(\ln x)) = f(g(h(x)))\)

\[\begin{align*} \text{Firstly, }& D(g\circ h) = \{x|x\in (0,\infty)\wedge \ln x\in (0,\infty)\} \\ & \ln x\in (0,\infty) \Leftrightarrow x\in (1,\infty) \\ \therefore \; \; & D(g\circ h) = x \in (1, \infty) \\ ~\\ \text{Now, }& D(f \circ (g\circ h)) = \{x|x\in (1,\infty)\wedge \ln(\ln x)\in (0,\infty)\} \\ & \ln(\ln x)\in (0,\infty) \Leftrightarrow \ln x\in (1,\infty) \Leftrightarrow x \in (e,\infty) \\ \therefore \; \; & D(f \circ g\circ h) = x \in (e, \infty) \end{align*}\]

\[\square\]

Example 2.1

Let \(f(x)=x+1\) and \(g(x)=x^2\) where \(D(g)=[-2,2]\).

Find \(D(f\circ g)\)

Solution:

\(\begin{align*} f\circ g(x) &= x^2+1 \\ D(f\circ g) &= \{x|x \in [-2,2] \wedge x^2 \in \mathbb{R} \} \\ & x^2 \in \mathbb{R} \Leftrightarrow x \in \mathbb{R} \\ \therefore \; \; & x \in [-2,2] \end{align*}\)

\[\square\]

Example 2.2

Consider the same constraints as in Example 2.1, but with \(D(f)=[-2,1]\)

Solution:

\(\begin{align*} D(f\circ g) &= \{x|x \in [-2,2] \wedge x^2 \in [-2,1] \} \\ & x^2 \in [-2,1] \Leftrightarrow x^2 \in [0,1] \Leftrightarrow x \in [-1,1] \\ \therefore \;\; & x \in [-1,1] \end{align*}\)

\[\square\]

References

Vaught, RL. 1995. Set theory: An introduction. 2nd edition. Boston: Birkhäuser.

 

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