Find Domain of Functions

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.1

Find domain of functions.

See attachment.

[Eigenvalue]

2 is always a set positive value, so x-1>0
x>1

[Eigenvalue]

66. [tex]\sqrt{x+3}[/tex]>0 because square roots cannot be negative for the result to be a real number
x+3>0
x>-3
However, x-2[tex]\ne[/tex]0, as dividing by 0 is undefined
x[tex]\ne[/tex]2

Domain: (-3,2) U (2, infinity)

[nycmath]

2 is always a set positive value, so x-1>0
x>1

This makes sense.

[nycmath]

66. [tex]\sqrt{x+3}[/tex]>0 because square roots cannot be negative for the result to be a real number
x+3>0
x>-3
However, x-2[tex]\ne[/tex]0, as dividing by 0 is undefined
x[tex]\ne[/tex]2

Domain: (-3,2) U (2, infinity)

Thank you for the data.