Given:
[tex]h(x)=\frac{6\sqrt[]{4x+3}}{9x-4}[/tex]a) The domain of the function is a set of all input values for which the function is real and defined.
First, find the on-negative values of radical and undefined points.
[tex]\begin{gathered} 4x+3\ge0\ldots\text{.non}-\text{negative values for radical} \\ 4x\ge-3 \\ x\ge-\frac{3}{4} \\ \text{the function is undefined when 9x-4=0} \\ 9x-4=0 \\ 9x=4 \\ x=\frac{4}{9} \end{gathered}[/tex]The domain is,
[tex]\: \lbrack-\frac{3}{4},\: \frac{4}{9})\cup(\frac{4}{9},\: \infty)[/tex]b) Range of the function is a set of dependant variables for which the function is defined.
The range is,
[tex]-\inftyAnswer:[tex]\text{Domain:}\lbrack-\frac{3}{4},\: \frac{4}{9})\cup(\frac{4}{9},\: \infty)[/tex]