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Math Help - reverse composition of functions......:(

  1. #1
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    Question reverse composition of functions......:(

    (a) If g(x) = 3x + 2 and h(x) = 9x2 + 12x + 9, find a function f such
    that f o g = h. (Think about what operations you would have to perform on the formula for g to end up with the formula for h.)
    f = 1
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    (b) If f(x) = 3x + 9 and h(x) = 3x2 + 3x + 6, find a function g such
    that f o g = h.
    g(x) = 2
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    how would i do such a question??
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  2. #2
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    Quote Originally Posted by Sneaky View Post
    (a) If g(x) = 3x + 2 and h(x) = 9x^2 + 12x + 9, find a function f such that f o g = h. (Think about what operations you would have to perform on the formula for g to end up with the formula for h.)

    f(x) = x^2 + 5

    __________________________


    (b) If f(x) = 3x + 9 and h(x) = 3x^2 + 3x + 6, find a function g such
    that f o g = h.

    g(x) = x^2 + x - 1
    for future reference, note that exponents can be designated by the caret symbol ...

    x^2 is x "squared"
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  3. #3
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    can u show me the exact steps into how you got that?
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    can u show me the exact steps into how you got that?
    no formal steps ... I just played with the given functions.

    for instance ...

    f(3x+2) = 9x^2 + 12x + 9

    if (3x+2) is squared, you get 9x^2 + 12x + 4

    note that you are short 5 units of the constant value. so ...

    f(x) = x^2 + 5


    for the second ...

    f[g(x)] = 3[g(x)] + 9 = 3x^2 + 3x + 6

    note that h(x) has coefficients of 3 for the first two terms ...

    so I tried 3(x^2 + x) + 9 = 3x^2 + 3x + 9

    but, the constant is 6 ... I have 3 units too many. How can I get rid of the additional 3 ?

    3(x^2 + x - 1) + 9 = 3x^2 + 3x - 3 + 9 = 3x^2 + 3x + 6
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