Answer:
[tex]y=\sqrt{x-2}[/tex]
Step-by-step explanation:
we know that
The domain of a function is the set of all possible values of x
Verify each case
case 1) we have
[tex]y=\sqrt[3]{x-2}[/tex]
The domain is the interval (-∞,∞)
therefore
The function is defined for all real numbers
case 2) we have
[tex]y=\sqrt{x-2}[/tex]
Remember that the radicand must be greater than or equal to zero
so
[tex]x-2\geq 0[/tex]
solve for x
[tex]x\geq 2[/tex]
The domain is the interval [2,∞)
therefore
The function is undefined for x=0
case 3) we have
[tex]y=\sqrt[3]{x+2}[/tex]
The domain is the interval (-∞,∞)
therefore
The function is defined for all real numbers
case 4) we have
[tex]y=\sqrt{x+2}[/tex]
Remember that the radicand must be greater than or equal to zero
so
[tex]x+2\geq 0[/tex]
solve for x
[tex]x\geq-2[/tex]
The domain is the interval [-2,∞)
The number zero is included in the domain
therefore
The function is defined for all real numbers greater than or equal to -2 (function is defined for x=0)
