For the given functions f and g, find the requested function and state its domain. Write the domain in interval notation.
-$f ( x ) = \frac { 8 x + 5 } { 2 x - 9 } ; g ( x ) = \frac { 4 x } { 2 x - 9 } ; \quad$ Find $f + g$.
A) $( f + g ) ( x ) = \frac { 12 x + 5 } { 2 x - 9 } ; ( * , 0 ) \cup ( 0 , \infty )$
B) $( f + g ) ( x ) = \frac { 12 x + 5 } { 2 x - 9 } ; \left( \infty , \frac { 9 } { 2 } \right) \cup \left( \frac { 9 } { 2 } , \infty \right)$
C) $( f + g ) ( x ) = \frac { 4 x - 5 } { 2 x - 9 } ; \left( \infty , \frac { 9 } { 2 } \right) \cup \left( \frac { 9 } { 2 } , \infty \right)$
D) $( f + g ) ( x ) = \frac { 12 x + 5 } { 2 x - 9 } ; \left( \infty , - \frac { 5 } { 12 } \right) \cup \left( - \frac { 5 } { 12 } , \infty \right)$

Question

Quizplus · Accepted Answer

To find $f + g$, we add the functions $f(x) = \frac{8x + 5}{2x - 9}$ and $g(x) = \frac{4x}{2x - 9}$ together. The common denominator is $2x - 9$, so we get $\frac{8x + 5 + 4x}{2x - 9} = \frac{12x + 5}{2x - 9}$. The domain excludes the value that makes the denominator zero, which is $x = \frac{9}{2}$, so the domain in interval notation is $\left( -\infty , \frac{9}{2} \right) \cup \left( \frac{9}{2} , \infty \right)$.

For the Given Functions F and G, Find the Requested f(x)=8x+52x−9;g(x)=4x2x−9;f ( x ) = \frac { 8 x + 5 } { 2 x - 9 } ; g ( x ) = \frac { 4 x } { 2 x - 9 } ; \quadf(x)=2x−98x+5​;g(x)=2x−94x​;

For the Given Functions F and G, Find the Requested $f ( x ) = \frac { 8 x + 5 } { 2 x - 9 } ; g ( x ) = \frac { 4 x } { 2 x - 9 } ; \quad$