- #1
murshid_islam
- 458
- 19
suppose x is a real variable. what is the domain of the function [tex]\sqrt{{x-3}\over{x-1}}[/tex]?
here [tex]{{x-3} \over {x-1}} \geq 0[/tex]
from this we get the domain to be [tex]\left(-\infty, 1\right) \cup \left[3, \infty\right)[/tex]
but if we write the function [tex]\sqrt{{x-3}\over{x-1}} = \frac{\sqrt{x-3}}{\sqrt{x-1}}[/tex]
then [tex]x-3 \geq 0 \Rightarrow x \geq 3[/tex]
and [tex]x-1 > 0 \Rightarrow x > 1[/tex]
from these two, we get the domain [tex]\left[3, \infty\right)[/tex]
why do i get two different domains?
here [tex]{{x-3} \over {x-1}} \geq 0[/tex]
from this we get the domain to be [tex]\left(-\infty, 1\right) \cup \left[3, \infty\right)[/tex]
but if we write the function [tex]\sqrt{{x-3}\over{x-1}} = \frac{\sqrt{x-3}}{\sqrt{x-1}}[/tex]
then [tex]x-3 \geq 0 \Rightarrow x \geq 3[/tex]
and [tex]x-1 > 0 \Rightarrow x > 1[/tex]
from these two, we get the domain [tex]\left[3, \infty\right)[/tex]
why do i get two different domains?
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