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    Tyan Bekooy's Avatar
    Tyan Bekooy Posts: 2, Reputation: 1
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    #1

    Dec 4, 2007, 11:48 AM
    Solving Composite Functions
    Evaluate gh(-4)

    g(x)= (4x+4)5
    h(x)= -1/3 (-x+5)

    I'm not getting the right answer.
    would any oneeeee help me ?

    xx
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #2

    Dec 4, 2007, 02:05 PM
    Can you show your work so far?
    Tyan Bekooy's Avatar
    Tyan Bekooy Posts: 2, Reputation: 1
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    #3

    Dec 4, 2007, 02:12 PM
    hmmm.
    apparently I do it kind of differently from the standard one,
    so I'm not sureee if you'll understand.
    but here it goes..

    g(x)= (4x+4)5
    h(x)= -1/3 (-x+5)
    = 1/3x -5/3

    gh(x)= g(1/3x -5/3)
    = 4 (1/3x - 5/3) +4 All Over 5
    = 4/3x - 20/3 + 4/1 All Over 5
    = 4/3x - 20/3 + 20/5 All over 5
    = 4/3x -20/3 - 20 All over 5
    gh (-4) = 4/3(-4) -20/3 -20 All over 5
    = -16/3 -20/3 -20 All over 5
    = -36/3 -20 All over 5
    = -12 - 20 Over 5
    = -36/5

    And the answer is supposed to be
    -1.6 or -8/5...
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #4

    Dec 5, 2007, 07:05 AM
    Thanks for showing your work! :)

    Since you want to solve for when x=-4 you don't need to do it the hard way and pull the x all the way through. Just substitute the -4 straight away and it will be much easier.



    Hope this helps!

    Now, your method will work also... but you have an error. I'm not sure why you converted the 4 into a 20/5, that was OK but of no use. But for some reason you then converted the 20/5 into a 20, that is your error. If you had left it as 4 or 20/5 you would have gotten 8/5.
    Elisha Grey's Avatar
    Elisha Grey Posts: 31, Reputation: 0
    Junior Member
     
    #5

    Jan 19, 2009, 03:29 PM

    h(-4) = -1/3(-(-4)+5) = -1/3(9) = -3
    gh(4) = g(-3) = (4(-3)+4)5 = (-8)5 = -40
    Samantha16's Avatar
    Samantha16 Posts: 1, Reputation: 1
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    #6

    Sep 21, 2010, 05:01 PM
    g(x)=3x/x+1
    Find 1) g2(x) 2)g3(x)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member
     
    #7

    Sep 22, 2010, 02:14 AM

    Hello Samantha16, it would be good if you started your own thread.



    Find
    1)
    2)



    Can you continue now? :)

    Do the same thing, but a little further for the second one. Post what you get! :)
    evie12's Avatar
    evie12 Posts: 1, Reputation: 1
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    #8

    Dec 23, 2010, 08:46 AM
    Hello, umm I've got an alevel question and wanted to know what the answer is but but can't find the mark scheme;

    f(x) = 2e(power x) domain minus infin. To infin.

    g(x) = 3lnx domain [1, infin.)

    a) Explain why gf(-1) does not exist?
    b) Find its simplest form an expression for fg(X). State the domain and range of fg.

    fg(X) = 2e(to the power of)3lnx
    = 2(3x)
    = 6x

    but don't know the domain and range or part a!


    Thanks :)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member
     
    #9

    Dec 23, 2010, 11:50 AM

    You could have posted a new thread for this...

    Anay, I wanted to confirm the question.





    And a) asks to show that does not exist?

    And b) asks to simplify fg(x)

    If so...

    a) Does exist it seems...



    Which is a straight line... in case it's the inverse, that too would exist.

    b) You didn't do it correctly, I'm afraid.



    And since this function usually exists from -infinity to +infinity, you have to take the smallest domain that you were given, that is [1, + infinity)

    The range is then calculated. What is fg(x) at x = 1. From there, you get your lowest value, until infinity. You range is thus from that value to positive infinity.

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