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Riggy · Answer 1 · 2023-01-13T12:52:49+0000

Given:

$h(x)=\frac{6\sqrt[]{4x+3}}{9x-4}$

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.

$\begin{gathered} 4x+3\ge0\ldots\text{.non}-\text{negative values for radical} \\ 4x\ge-3 \\ x\ge-(3)/(4) \\ \text{the function is undefined when 9x-4=0} \\ 9x-4=0 \\ 9x=4 \\ x=(4)/(9) \end{gathered}$

The domain is,

$\: \lbrack-(3)/(4),\: (4)/(9))\cup((4)/(9),\: \infty)$

b) Range of the function is a set of dependant variables for which the function is defined.

The range is,

$-\infty<strong>Answer:</strong>[tex]\text{Domain:}\lbrack-(3)/(4),\: (4)/(9))\cup((4)/(9),\: \infty)$

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