Example 3.4. Algebraic Constraints on Domain.
Find the domain of each algebraic expression
- \(\displaystyle \displaystyle 4-x\)
- \(\displaystyle \displaystyle \sqrt{4-x}\)
- \(\displaystyle \displaystyle \frac{1}{\sqrt{4-x}}\)
- \(\displaystyle \displaystyle \frac{1}{\sqrt{4+x^2}}\)
- \(\displaystyle \displaystyle \frac{1}{x-y}\)
Solution.
- 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 \((-\infty,+\infty)\text{.}\)
- 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 \((-\infty,4]\text{.}\)
- Additionally, we require that we never divide by zero. This requires \(4-x \gt 0\) and the domain is now the open interval \((-\infty,4)\text{.}\)
- 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 \(\mathbb{R} = (-\infty,+\infty)\)
- We require that \(x-y \neq 0\text{.}\) This means \(x \neq y\text{.}\) So the domain is the set of all points \((x,y)\) in the coordinate plane with \(x\neq y\text{.}\)
