Given the function:
[tex]h(x)=\frac{6\sqrt[]{4x+3}}{9x-4}[/tex]
Let's find the domain of the function.
The domain of a function is the possible values of x which makes the function true.
To find the domain of the function, set the denominator to zero and solve for x.
We have:
9x - 4 = 0
Add 4 to both sides:
9x - 4 + 4 = 0 + 4
9x = 4
Divide both sides of the equation by 9:
[tex]\begin{gathered} \frac{9x}{9}=\frac{4}{9} \\ \\ x=\frac{4}{9} \end{gathered}[/tex]
Also, set the expression in the radicand greater or equal to zero and solve for x.
4x + 3 ≥ 0
Subtract 3 from both sides:
4x + 3 - 3 ≥ 0 - 3
4x ≥ -3
Divide both sides by 4:
[tex]\begin{gathered} \frac{4x}{4}\ge-\frac{3}{4} \\ \\ x\ge-\frac{3}{4} \end{gathered}[/tex]
The domain of h is:
[tex]\mleft\lbrace x\vert x\ge-\frac{3}{4},x\ne\frac{4}{9}\mright\rbrace[/tex]
Therefore, the domain of h in interval notation is:
[tex]\lbrack-\frac{3}{4},\frac{4}{9})\cup(\frac{4}{9},\infty)[/tex]
