Answer
$\color{blue}{\bf{ -7x + 7 }; \text{ domain } (-\infty,\infty) }$
$\color{blue}{\bf{ x+5 }; \text{ domain } (-\infty,\infty) }$
$\color{blue}{\bf{ 12x^2-27x+6 }; \text{ domain } (-\infty,\infty) }$
$\color{blue}{\bf{ \dfrac{ 6 - 3x }{ -4x + 1 } }; \text{ domain } (-\infty, \dfrac{1}{4} )\bigcup( \dfrac{1}{4} ,\infty) }$
Work Step by Step
We are given the two functions $\bf{f}$ and $\bf{g}$
$\bf{f(x) = 6 - 3x }$ and $\bf{g(x) = -4x + 1 }$
We are asked to find $\bf{( f+g )( x )}$ and its domain:
$( 6 - 3x )+( -4x + 1 ) $
$- 3x -4x + 6+1 $
$\color{blue}{\bf{ -7x + 7 }}$
$x$ can be any real number so its domain is:
$\color{blue}{\bf{ (-\infty,\infty) }}$
We are asked to find $\bf{( f-g )( x )}$ and its domain:
$( 6 - 3x )-( -4x + 1 )$
$ 6 - 3x +4x - 1 $
$ 4x - 3x +6 - 1 $
$\color{blue}{\bf{ x+5 }}$
$x$ can be any real number so its domain is:
$\color{blue}{\bf{ (-\infty,\infty) }}$
We are asked to find $\bf{( fg )( x )}$ and its domain:
$( 6 - 3x )( -4x + 1 ) $
$-24x+6+12x^2-3x$
$\color{blue}{\bf{ 12x^2-27x+6 }}$
$x$ can be any real number so its domain is:
$\color{blue}{\bf{ (-\infty,\infty) }}$
We are asked to find $\bf{( \dfrac{f }{ g} )( x)}$ and its domain:
$\color{blue}{\bf{ \dfrac{ 6 - 3x }{ -4x + 1 } }}$
Since division by $0$ is undefined,
$ -4x + 1 \neq 0 $
$ -4x \neq -1 $
$x \neq \dfrac{1}{4} $ so its domain is:
$\color{blue}{\bf{ (-\infty, \dfrac{1}{4} )\bigcup( \dfrac{1}{4} ,\infty) }}$
