If [tex]f(x)=\sqrt{4x+9}+2[/tex], which inequality can be used to find the domain of [tex]f(x)[/tex]?

A. [tex]\sqrt{4x} \geq 0[/tex]
B. [tex]4x + 9 \geq 0[/tex]
C. [tex]4x \geq 0[/tex]
D. [tex]\sqrt{4x + 9} + 2 \geq 0[/tex]



Answer :

To find the domain of [tex]\( f(x) = \sqrt{4x + 9} + 2 \)[/tex], we need to ensure that the expression under the square root is non-negative. The square root function, [tex]\(\sqrt{y}\)[/tex], is only defined for [tex]\( y \geq 0 \)[/tex].

Let's start by examining the expression under the square root in the function [tex]\( f(x) \)[/tex]:
[tex]\[ \sqrt{4x + 9} \][/tex]

For this expression to be defined, the expression inside the square root must be non-negative:
[tex]\[ 4x + 9 \geq 0 \][/tex]

This is the inequality that ensures the square root is defined. Solving this inequality for [tex]\( x \)[/tex]:

1. First, isolate [tex]\( x \)[/tex] by subtracting 9 from both sides:
[tex]\[ 4x \geq -9 \][/tex]

2. Then, divide both sides by 4:
[tex]\[ x \geq -\frac{9}{4} \][/tex]

Therefore, the inequality that can be used to find the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ 4x + 9 \geq 0 \][/tex]

Thus, among the given options, the correct inequality is:
[tex]\[ 4 x + 9 \geq 0 \][/tex]

This inequality ensures that the expression under the square root is non-negative, and hence, the function [tex]\( f(x) = \sqrt{4x + 9} + 2 \)[/tex] is well-defined.