Answer:
x - 3 ≥ 0
Step-by-step explanation:
The given function is f(x) = [tex]\sqrt{x-3}[/tex]
Here we need to find the inequality that is used to find the domain of the function.
We should not have negative number inside the square root because if negative number comes inside the square root, then it is not a function.
Therefore, (x - 3) must be greater than or equal to 0
That is x - 3 ≥ 0
We use the above inequality to find the domain of the function.
When we solve the above inequality, we get the domain of the function.
x - 3≥0
add 3 on both sides, we get
x - 3 + 3 ≥ 0 + 3
x ≥ 3
Domain is the set of all real numbers greater than or equal 3.
Therefore, the answer to the question is x - 3 ≥ 0
Answer:
[tex]x-3\geq 0[/tex]
Step-by-step explanation:
We have been given a function [tex]f(x)=\sqrt{x-3}[/tex]. We are asked to find the inequality that can be used to find the domain of f(x).
We know that a square root function is defined for non-negative values of independent variable.
To find the domain of our given function, we will set the expression under radical greater than or equal to 0 as:
[tex]x-3\geq 0[/tex]
Therefore, our required inequality would be [tex]x-3\geq 0[/tex].
Let us find domain of the function.
[tex]x-3+3\geq 0+3[/tex]
[tex]x\geq 3[/tex]
Therefore, the domain of the function is all values of x greater than or equal to 3 that is [tex][3,\infty)[/tex] in interval notation.
