What are the domain and range of the function [tex]$f(x)=\sqrt{x-7}+9$[/tex]?

A. Domain: [tex]x \geq -7[/tex], Range: [tex]y \geq 9[/tex]
B. Domain: [tex]x \geq 7[/tex], Range: [tex]y \geq -9[/tex]
C. Domain: [tex]x \geq 7[/tex], Range: [tex]y \geq 9[/tex]
D. Domain: [tex]x \geq 9[/tex], Range: [tex]y \geq 7[/tex]



Answer :

To determine the domain and range of the function [tex]\( f(x) = \sqrt{x - 7} + 9 \)[/tex], let's go through the problem step-by-step.

### Domain

1. Identify the condition for the square root function:

The square root function, [tex]\(\sqrt{x - 7}\)[/tex], is defined only for non-negative arguments. Hence,
[tex]\[ x - 7 \geq 0 \][/tex]

2. Solve for [tex]\( x \)[/tex]:

[tex]\[ x \geq 7 \][/tex]

Therefore, the domain of [tex]\( f(x) = \sqrt{x - 7} + 9 \)[/tex] is:

[tex]\[ \boxed{x \geq 7} \][/tex]

### Range

1. Identify the minimum value of [tex]\( f(x) \)[/tex]:

When [tex]\( x = 7 \)[/tex],
[tex]\[ f(7) = \sqrt{7 - 7} + 9 = \sqrt{0} + 9 = 9 \][/tex]

2. Behavior of the function:

As [tex]\( x \)[/tex] increases beyond 7, [tex]\(\sqrt{x - 7}\)[/tex] becomes positive and increases. Therefore, the function [tex]\( f(x) = \sqrt{x - 7} + 9 \)[/tex] increases without bound.

3. Determine the range:

Since the smallest value of [tex]\(\sqrt{x - 7}\)[/tex] is 0 (when [tex]\( x = 7 \)[/tex]) and increases thereafter, the smallest value of [tex]\( f(x) \)[/tex] is 9, increasing as [tex]\( x \)[/tex] increases.

Thus, the range of [tex]\( f(x) \)[/tex] is:

[tex]\[ \boxed{y \geq 9} \][/tex]

Given these domain and range findings, the correct choice from the provided options is:

- Domain: [tex]\( x \geq 7 \)[/tex]
- Range: [tex]\( y \geq 9 \)[/tex]

So, the correct answer is:

[tex]\[ \boxed{3} \][/tex]