Answer :
To find the domain of the function [tex]\( h'(x) = \sqrt{x-7} + 5 \)[/tex], we need to ensure that all parts of the function are defined for real numbers.
The square root function, [tex]\(\sqrt{x}\)[/tex], is only defined for values where the expression under the square root is non-negative. This means the expression inside the square root, in this case, [tex]\(x-7\)[/tex], must be greater than or equal to zero.
Let's find out when [tex]\(x-7\)[/tex] is non-negative:
1. Set up the inequality:
[tex]\[
x - 7 \geq 0
\][/tex]
2. Solve for [tex]\(x\)[/tex]:
[tex]\[
x \geq 7
\][/tex]
This inequality tells us that [tex]\(x\)[/tex] must be greater than or equal to 7 for the function [tex]\( h'(x) = \sqrt{x-7} + 5 \)[/tex] to be defined for real numbers.
Therefore, the domain of the function [tex]\(h'(x)\)[/tex] is [tex]\(x \geq 7\)[/tex].
The correct answer is:
A. [tex]\(x \geq 7\)[/tex]
The square root function, [tex]\(\sqrt{x}\)[/tex], is only defined for values where the expression under the square root is non-negative. This means the expression inside the square root, in this case, [tex]\(x-7\)[/tex], must be greater than or equal to zero.
Let's find out when [tex]\(x-7\)[/tex] is non-negative:
1. Set up the inequality:
[tex]\[
x - 7 \geq 0
\][/tex]
2. Solve for [tex]\(x\)[/tex]:
[tex]\[
x \geq 7
\][/tex]
This inequality tells us that [tex]\(x\)[/tex] must be greater than or equal to 7 for the function [tex]\( h'(x) = \sqrt{x-7} + 5 \)[/tex] to be defined for real numbers.
Therefore, the domain of the function [tex]\(h'(x)\)[/tex] is [tex]\(x \geq 7\)[/tex].
The correct answer is:
A. [tex]\(x \geq 7\)[/tex]
