Answer
$\color{blue}{\bf{ 5x - 1 }; \text{ domain } (-\infty,\infty) }$
$\color{blue}{\bf{ x +9 }; \text{ domain } (-\infty,\infty) }$
$\color{blue}{\bf{ 6x^2-7x-20 }; \text{ domain } (-\infty,\infty) }$
$\color{blue}{\bf{ \dfrac{ 3x + 4 }{ 2x - 5 } }; \text{ domain } (-\infty,\dfrac{5}{2})\bigcup(\dfrac{5}{2},\infty) }$
Work Step by Step
We are given the two functions $\bf{f}$ and $\bf{g}$
$\bf{f(x) = 3x + 4 }$ and $\bf{g(x) = 2x - 5 }$
We are asked to find $\bf{( f+g )( x )}$ and its domain:
$3x + 4+ 2x - 5 $
$\color{blue}{\bf{ 5x - 1 }}$
$x$ 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:
$3x + 4 -(2x - 5)$
$3x -2x +4+ 5$
$\color{blue}{\bf{ x +9 }}$
$x$ 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:
$(3x + 4)(2x - 5) $
$6x^2-15x+8x-20$
$\color{blue}{\bf{ 6x^2-7x-20 }}$
$x$ 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:
$\color{blue}{\bf{ \dfrac{ 3x + 4 }{ 2x - 5 } }}$
Since division by $0$ is undefined,
$2x - 5\neq0$
$2x \neq5$
$x \neq\dfrac{5}{2}$ so its domain is:
$\color{blue}{\bf{ (-\infty,\dfrac{5}{2})\bigcup(\dfrac{5}{2},\infty) }}$
