# Math Help - Domain of composite function

1. ## Domain of composite function

I need help finding the domain of the composite function f*g

1. f(x)= x/x+6, g(x)= 12/x+3

2. f(x)= 4x+24, g(x)= x+9

I also need help finding the domain for

f-g
3. f(x)=6x-3 g(x)= 4x-5

2. ## Re: Domain of composite function

1) f*g(x)
=f(12/x+3)
=(12/x+3)/(12/x+3)+6
=6
Domain of composite function f*g = domain of g = (minus infinity to plus infinity) exclude x=0

3. ## Re: Domain of composite function

2) f*g
=f(x+9)
=4(x+9)+24
=4x+60
Domain of composite function f*g = domain of g = (minus infinity to plus infinity)

4. ## Re: Domain of composite function

3) f-g
=2x+2
Domain of function f - g = minus infinity to plus infinity (I guess this is the answer though I am not very certain)

5. ## Re: Domain of composite function

Hello, Lilmissmimiboots!

Find the domain of the composite function $f\circ g.$

$1.\;\;f(x)= \frac{x}{x+6},\;\;g(x)= \frac{12}{x+3}$

$f\circ g \;=\;f(g(x)) \;=\;f\left(\tfrac{12}{x+3}\right) \;=\;\frac{\frac{12}{x+3}}{\frac{12}{x+3}+6}$

Multiply by $\tfrac{x+3}{x+3}\!:$

. . $f\circ g \;=\;\frac{(x+3)\left(\frac{12}{x+3}\right)}{(x+3) \left(\frac{12}{x+3} + 6\right)} \;=\;\frac{12}{12 + 6(x+3)}$

Hence: . $f\circ g \;=\;\frac{12}{6x+30} \quad\Rightarrow\quad f\circ g \;=\;\frac{2}{x+5}$

Domain: . $(\text{-}\infty,\text{-}5) \cup (\text{-}5, \infty)$

Find the domain of $f\circ g.$

$2.\;\;f(x)= 4x+24,\;\; g(x)= x+9$

$f\circ g \;=\;f(x + 9) \;=\;4(x+9) + 24 \;=\;4x + 60$

Domain: . $(\text{-}\infty,\infty)$

Find the domain of $f-g.$

$3.\;\;f(x)=6x-3 ,\;\;g(x)= 4x-5$

$f-g \;=\;(6x-3) - (4x -5) \;=\;2x+2$

Domain: . $(\text{-}\infty,\,\infty)$

6. ## Re: Domain of composite function

Originally Posted by Soroban
Hello, Lilmissmimiboots!

$f\circ g \;=\;f(g(x)) \;=\;f\left(\tfrac{12}{x+3}\right) \;=\;\frac{\frac{12}{x+3}}{\frac{12}{x+3}+6}$

Multiply by $\tfrac{x+3}{x+3}\!:$

. . $f\circ g \;=\;\frac{(x+3)\left(\frac{12}{x+3}\right)}{(x+3) \left(\frac{12}{x+3} + 6\right)} \;=\;\frac{12}{12 + 6(x+3)}$

Hence: . $f\circ g \;=\;\frac{12}{6x+30} \quad\Rightarrow\quad f\circ g \;=\;\frac{2}{x+5}$

Domain: . $(\text{-}\infty,\text{-}5) \cup (\text{-}5, \infty)$
Not quite. If $x=-3$, then $g(x)$ is not defined, so $(f\circ g)(-3) = f(g(-3))$ is not defined, either. Hence, the domain is
$(-\infty,-5)\cup(-5,-3)\cup(-3,\infty)$.