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Math Help - composition of funtion

  1. #1
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    Unhappy composition of funtion

    a little confuse with this question

    Find(fogoh)^-1

    I dont know How my teacher get it ash^-1(g^-1(f^-1(x)))

    I thought it would bef^-1(g^-1(h^-1(x)))

    Because if i have Just(fog)^-1(x)i would getf^-1(g(x))

    I HAD to find f^-1(x) , g^-1(x) , h^-1(x) along with (fogoh)^-1

    F(X)= 3x+4 answer for this is x-4/3
    g(x) = 2/x answer is 2/x
    h(x)= radical x-4 answer is x^2+4

    I know how to get those i just dont understand (fogoh)^-1 and how
    To put it together

    He put them together as



    h^-1(g^-1(x-4/3))=h^-1(2/(x-4/3)) = h^-1(6/x-4) = (6/x-4)^2 +4

    I just dont understnad the (fogoh)^-1 part can u please help me with this

    Thank you.
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  2. #2
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    Re: composition of funtion

    Quote Originally Posted by Candy101 View Post
    a little confuse with this question
    Find(fogoh)^-1
    I dont know How my teacher get it ash^-1(g^-1(f^-1(x)))
    Do you understand that f^{-1}\circ f=x~?

    Then (h^{-1}\circ g^{-1}\circ f^{-1})\circ(f\circ g\circ h)=~?

    Work from inside out.
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  3. #3
    Behold, the power of SARDINES!
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    Re: composition of funtion

    If you draw a diagram of the domains and ranges it will make the situation clearer on why the order must be reversed.

    By the way, in general, this is called the reverse order law.

    Rember that the domain of f is the range of f inverse and the range of f is the domian of f inverse.

    composition of funtion-capture.png
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  4. #4
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    Re: composition of funtion

    The "work from inside out" concpet help me to remember it
    But im having a lil trouble putting it together.

    F^-1= x-4/3
    g^-1=2/x
    h^-1= x^2+4

    h^-1(g^-1(f^-1(x)))

    I replace (F^-1) (x) with x-4/3


    h^-1(g^-1( x-4/3 ))

    Then i use 2/x and for g^-1 and i replace all the x in 2/x witb x-4/3

    So this is what i got so far

    h^-1(2/(x-4/3))

    And the teacher got the same thing
    But then got
    h^-1 (6/x-4)= (6/x-4)^2+4 i dont understsnd where he got the 6 from. How did he get (6/x-4)= from h^-1(2/(x-4/3))?
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  5. #5
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    Re: composition of funtion

    Never mind i got it! Thanks guys for all the help!!!
    I can actually do these kind of problems thanks to you!
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  6. #6
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    Re: composition of funtion

    Hello, Candy101!

    Everyone (including your teacher) is making very hard work . . .


    \text{Given: }\:\begin{Bmatrix} f(x) &=& 3x + 4 \\ \\[-4mm] g(x) &=& \dfrac{2}{x} \\ \\[-3mm] h(x) &=& \sqrt{x-4} \end{Bmatrix}

    \text{Find }\,(f\circ g\circ h)^{-1}

    First find f\circ g\circ h, then find its inverse . . .


    f\left(g\big[h(x)\big]\right) \;=\;f\left(g\bog[\sqrt{x-4}\big]\right) \;=\; f\left(\frac{2}{\sqrt{x-4}}\right) \;=\;3\cdot\frac{2}{\sqrt{x-4}} + 4

    We have: . y \;=\;\frac{6}{\sqrt{x-4}} + 4


    To find the inverse function:
    . . Switch x and y; solve for y.

    We have: . x \;=\;\frac{6}{\sqrt{y-4}} + 4

    . . . . . x - 4 \:=\:\frac{6}{\sqrt{y-4}}

    . . . \sqrt{y-4} \;=\;\frac{6}{x-4}

    . . . . y - 4 \;=\;\left(\frac{6}{x-4}\right)^2

    n . . . . . y \;=\;\left(\frac{6}{x-4}\right)^2 + 4


    Therefore: . \big(f\circ g\circ h\big)^{-1} \;=\;\left(\frac{6}{x-4}\right)^2 + 4
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