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Math Help - Domain of composite function

  1. #1
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    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
    Last edited by Lilmissmimiboots; April 15th 2014 at 01:41 PM.
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  2. #2
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    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
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  3. #3
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    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)
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  4. #4
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    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)
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  5. #5
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    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)
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  6. #6
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    Re: Domain of composite function

    Quote Originally Posted by Soroban View Post
    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).
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