Answers
Given:
The functions are:
(1) [tex]h(x)=\sqrt{1-x}[/tex]
(2) [tex]x+y=4[/tex]
(3) [tex]x^2+y^2=16[/tex]
To find:
The domain of given functions.
Solution:
Domain is the set of input and x -values.
Value under the square root must be positive because if the number under square root is negative, then it is a complex or imaginary number.
(1) [tex]h(x)=\sqrt{1-x}[/tex]
The function is defined if
[tex]1-x\geq 0[/tex]
[tex]1\geq x[/tex]
The function is defined for all real values of x which are less than or equal to 1.
Therefore, the domain of the function is (-∞,1].
(2) [tex]x+y=4[/tex]
It is a linear function and linear function is defined for all real values of x.
Therefore, the domain of the function is (-∞,∞).
(3) [tex]x^2+y^2=16[/tex]
It can be written as
[tex]y^2=16-x^2[/tex]
[tex]y=\pm \sqrt{16-x^2}[/tex]
It is defined if
[tex]16-x^2\geq 0[/tex]
[tex]16\geq x^2[/tex]
[tex]4\geq x\geq -4[/tex]
This function defined for all real values of x from -4 to 4.
Therefore, the domain of the function is [-4,4].
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