Answer
$\color{blue}{\bf{ \sqrt{4x-1} + \dfrac{1}{x} }}$
$\color{blue}{\bf{ \sqrt{4x-1} - \dfrac{1}{x} } }$
$\color{blue}{\bf{ \dfrac{ \sqrt{4x-1} }{x} } }$
$\color{blue}{\bf{ x\sqrt{4x-1} }}$
$\color{blue}{\bf{ \text{ all domains are } [ \dfrac{1}{4} ,\infty) }}$
Work Step by Step
We are given the two functions $\bf{f}$ and $\bf{g}$
$\bf{f(x) = \sqrt{4x-1} }$ and $\bf{g(x) = \dfrac{1}{x} }$
We are asked to find $\bf{( f+g )( x )}$ and its domain.
$f(x)+g(x)$
$\color{blue}{\bf{ \sqrt{4x-1} + \dfrac{1}{x} }}$
$x\neq0$ because $\dfrac{1}{0}$ is undefined
$4x-1\geq0$ or $ \sqrt{4x-1}$ would not be a real number
$4x\geq1$
$x\geq\dfrac{1}{4}$
Therefore the domain is: $\color{blue}{\bf{[ \dfrac{1}{4} ,\infty) }}$
We are asked to find $\bf{( f-g )( x )}$ and its domain.
$f(x)-g(x)$
$\color{blue}{\bf{ \sqrt{4x-1} - \dfrac{1}{x} }}$
Its domain is the same as above, $\color{blue}{\bf{[ \dfrac{1}{4} ,\infty) }}$
We are asked to find $\bf{( fg )( x )}$ and its domain.
$( \sqrt{4x-1} )( \dfrac{1}{x} ) $
$\color{blue}{\bf{ \dfrac{ \sqrt{4x-1} }{x} }}$
$x$ can be any real number so its domain is:
$\color{blue}{\bf{ (-\infty,\infty) }}$
Its domain is the same as above, $\color{blue}{\bf{[ \dfrac{1}{4} ,\infty) }}$
We are asked to find $\bf{( \dfrac{f }{ g} )( x)}$ and its domain.
$\dfrac{ \sqrt{4x-1} }{ \dfrac{1}{x} }$
$ \dfrac{ \sqrt{4x-1} }{1}\div\dfrac{1}{x}$
$ \dfrac{ \sqrt{4x-1} }{1}\times{\dfrac{x}{1}}$
$\color{blue}{\bf{ x\sqrt{4x-1} }}$
Its domain is the same as above, $\color{blue}{\bf{[ \dfrac{1}{4} ,\infty) }}$
