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Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\documentclass{article}
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\title{The domain of a composite function}
\author{Aidan Horn}
\begin{document}
\maketitle
I have tried to find a systematic way of finding the domain of a composite function for a few months now. Today, I tried explaining this concept to a first-year maths student, and found myself horribly confused all over again. After some research in the library, I have finally found a definition of that can be used for this purpose (the internet mostly directs one to worked-through examples and no generalized definition, so I had to resort to the library):
$$D(f \circ g) = \{x|x\in D(g) \wedge g(x) \in D(f)\}$$
\raggedleft{(Vaught, 1995:18)}
\raggedright{
Where $D(\lambda) = $ the domain of $\lambda$
\section*{Example 1} ~\\
Solve $$D(\ln(\ln(\ln x)))$$ (This was in a MAM1000W past paper)
~\\
\underline{Solution:}
Let $\ln(\ln(\ln x)) = f(g(h(x)))$
\begin{minipage}{3in}
\begin{align*}
& D(g\circ h) = \{x|x\in (0,\infty)\wedge \ln x\in (0,\infty)\} \\
\text{Now, } & \ln x\in (0,\infty) \Leftrightarrow x\in (1,\infty) \\
\therefore \; \; & x \in (1, \infty) \\
~\\
& D(f \circ (g\circ h)) = \{x|x\in (1,\infty)\wedge \ln(\ln x)\in (0,\infty)\} \\
\text{Now, } & \ln(\ln x)\in (0,\infty) \Leftrightarrow \ln x\in (1,\infty) \Leftrightarrow x \in (e,\infty) \\
\therefore \; \; & x \in (e, \infty)
\end{align*}
\end{minipage}
}
~\\
\raggedleft{$\blacksquare$}
\raggedright{
\section*{Example 2.1}
Consider $f(x)=x+1$ and $g(x)=x^2$ where $D(g)=[-2,2]$.
Find $D(f\circ g)$
~\\
\begin{minipage}{2in}
\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*}
\end{minipage}
}
~\\
\raggedleft{$\blacksquare$}
\raggedright{
\section*{Example 2.2}
Consider the same example as Example 2.1, but with $D(f)=[-2,1]$
\begin{minipage}{3in}
\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*}
\end{minipage}
}
~\\
\raggedleft{$\blacksquare$}
\raggedright{
\pagebreak
\section*{References}
\hangindent=0.7cm
Vaught, RL. 1995. Set theory: An introduction. 2nd edition. Boston: Birkh$\ddot{a}$user.
}
\end{document}