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Math Help - composite functions 2

  1. #1
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    composite functions 2

    Hello everyone,
    Here are some questions.I dont have the answers to them and im kind of having trouble. I would really appreciate it if you showed me the way of solving A and just the answers to the rest because i want to solve them myself.

    Given that f(x)= 5-x and g(x)= 4/x , where x is all real numbers except for 0 and 5 find the value

    a) gg(1)
    b) fffff(6)
    c) fggf(2)

    Thank you
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  2. #2
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    Quote Originally Posted by Oasis1993 View Post
    Hello everyone,
    Here are some questions.I dont have the answers to them and im kind of having trouble. I would really appreciate it if you showed me the way of solving A and just the answers to the rest because i want to solve them myself.

    Given that f(x)= 5-x and g(x)= 4/x , where x is all real numbers except for 0 and 5 find the value

    a) gg(1)
    b) fffff(6)
    c) fggf(2)

    Thank you
    g(x) = 4/x

    g(1) = 4/1 = 4

    g(4) = 4/4 = 1

    so ...

    g[g(1)] = g[4] = 1
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  3. #3
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    thank you.
    and what about b?
    would i have to insert 6, 5 times?
    I dont really get how to?
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  4. #4
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    Quote Originally Posted by Oasis1993 View Post
    thank you.
    and what about b?
    would i have to insert 6, 5 times? no, that's not how it works.
    f(x) = 5-x

    note that f(6) = 5 - 6 = -1 and f(-1) = 5 - (-1) = 6

    work inside out ...

    f[f[f[f[f(6)] =

    f[f[f[f(-1)] =

    f[f[f(6)] =

    f[f(-1)] =

    f(6) = -1

    get the idea?
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    Quote Originally Posted by Oasis1993 View Post
    Hello everyone,
    Here are some questions.I dont have the answers to them and im kind of having trouble. I would really appreciate it if you showed me the way of solving A and just the answers to the rest because i want to solve them myself.

    Given that f(x)= 5-x and g(x)= 4/x , where x is all real numbers except for 0 and 5 find the value

    a) gg(1)
    b) fffff(6)
    c) fggf(2)

    Thank you
    f(x) = 5-x
    ff(x) = 5-(5-x) = x
    fff(x) = 5-x

    Note how this repeats itself. If there is an odd number of f then it will be 5-x. Even numbers will be x.

    fffff(x) = 5-x
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  6. #6
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    Thank you very much!

    so for question C

    fggf(2)

    would it be like this?

    f(2)= 5-2=3
    g(3)=4/3
    g(4/3)=4/(4/3)
    f(3.08)=5-3.08= 1.92

    ?
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  7. #7
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    Quote Originally Posted by Oasis1993 View Post
    Thank you very much!

    so for question C

    fggf(2)

    would it be like this?

    f(2)= 5-2=3
    g(3)=4/3
    g(4/3)=4/(4/3) e^(i*pi) = \frac{4}{\frac{4}{3}} = 4 \times \frac{3}{4} = 3

    f(3.08)=5-3.08= 1.92

    ?
    g^{-1}(x) = \frac{4}{x} = g(x)

    Therefore wherever you get two lots of g next to each other they will cancel to leave x.

    fggf(x) = ff(x) = x as established in question 2.

    Therefore fggf(2) = 2


    The function h(x) = \frac{a}{x} \: , \: a,x \neq 0 is equal to it's inverse
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  8. #8
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    Quote Originally Posted by Oasis1993 View Post
    Thank you very much!

    so for question C

    fggf(2)

    would it be like this?

    f(2)= 5-2=3
    g(3)=4/3
    g(4/3)=4/(4/3)
    f(3.08)=5-3.08= 1.92 ... no

    ?
    put away the calculator ...

    4/(4/3) = 4(3/4) = 3

    f(3) = 2
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  9. #9
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    Thank you!!
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