Example 4.2.1. Algebraic Constraints on Domain.
Find the domain of each algebraic expression
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\(\displaystyle \displaystyle 4-x\)
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\(\displaystyle \displaystyle \frac{1}{4-x}\)
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\(\displaystyle \displaystyle \sqrt{4-x}\)
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\(\displaystyle \displaystyle \frac{1}{\sqrt{4-x}}\)
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\(\displaystyle \displaystyle \frac{1}{\sqrt{4+x^2}}\)
Solution.
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The operation of subtracting \(x\) from \(4\) may be performed to any real number. Thus, the domain is the set of all real numbers which we could write as \(\mathbb{R}\) or in interval notation as \((-\infty,+\infty)\text{.}\)
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We must never divide by zero! (See Warning 2.1.2.) So the domain consists of all real numbers \(x\text{,}\) except \(x=4\text{.}\) This may be written succinctly as “all real numbers \(x\neq 4\)”; or, in interval notation as the union of the intervals on either side of \(x=4\text{:}\)\begin{equation*} (-\infty,4) \cup (4,\infty). \end{equation*}Recall from Section 2.9, that a real number is in this set if it is less than \(4\) OR if it is greater than \(4\text{.}\)
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Since a square root is now involved, we must require \(4-x \geq 0\text{.}\) Equivalently, \(4 \geq x\text{.}\) Thus, the domain is the interval \((-\infty,4]\text{.}\)
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Additionally, we require that we never divide by zero. This requires \(4-x \gt 0\) and the domain is now the open interval \((-\infty,4)\text{.}\)
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As before, we require that \(4+x^2 \gt 0\text{.}\) But squaring any real number results in a non-negative quantity and adding \(4\) will guarantee the result is positive. Thus, the domain is the set of all real numbers \(\mathbb{R} = (-\infty,+\infty)\)
