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

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

Sagot :

To determine the domain of the function \( f(x) = \sqrt{4x + 9} + 2 \), we need to ensure that the expression inside the square root is non-negative. This is because the square root of a negative number is not defined within the set of real numbers.

To do so, we solve the inequality:

[tex]\[ 4x + 9 \geq 0 \][/tex]

Let’s break down the solution step by step.

1. Write the inequality:
We need the term under the square root to be greater than or equal to zero.
[tex]\[ 4x + 9 \geq 0 \][/tex]

2. Isolate \( x \):
Subtract 9 from both sides of the inequality.
[tex]\[ 4x \geq -9 \][/tex]

3. Solve for \( x \):
Divide both sides by 4 to isolate \( x \).
[tex]\[ x \geq -\frac{9}{4} \][/tex]

Therefore, the inequality \( 4x + 9 \geq 0 \) allows us to determine the domain of the function. Hence, the correct inequality to use is:

[tex]\[ 4 x + 9 \geq 0 \][/tex]