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    Hi,
    So we're going back to school on Monday and I have some work to do before we go back but I'm a bit rusty because we've just had summer! If anyone can help me out / talk through part 3a 3b and 3c of this question it would be appreciated! Name:  ImageUploadedByStudent Room1472833028.290883.jpg
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    (Original post by james0902)
    Hi,
    So we're going back to school on Monday and I have some work to do before we go back but I'm a bit rusty because we've just had summer! If anyone can help me out / talk through part 3a 3b and 3c of this question it would be appreciated! Name:  ImageUploadedByStudent Room1472833028.290883.jpg
Views: 146
Size:  135.2 KB


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    3a : What is fg(a) ?

    Once you've found this, set it equal to 100 to find a.
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    EDIT - Help with question 4 would also be appreciated.


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    Remember, composite function fg(x) is the same as f(g(x)) - i.e. substituting the function g(x) as the variable x in the function f(x).

    E.g.

    If f(x) = x - 2 and g(x) = 3x, the function fg(x) = 3x - 2.

    So if we had to find the value of a for which fg(a) = 7, we get 3a - 2 = 7.

    a = 3
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    (Original post by notnek)
    3a : What is fg(a) ?

    Once you've found this, set it equal to 100 to find a.
    Name:  ImageUploadedByStudent Room1472835023.369152.jpg
Views: 150
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    So this is where I'm at with the working I've put the 100 over to this side but I'm just struggling with solving this! Any help is appreciated (for question 3a)


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    (Original post by james0902)
    Name:  ImageUploadedByStudent Room1472835023.369152.jpg
Views: 150
Size:  97.0 KB

    So this is where I'm at with the working I've put the 100 over to this side but I'm just struggling with solving this! Any help is appreciated (for question 3a)


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    fg(x) doesn't mean f(x) \times g(x).

    It means f(g(x)).

    If you're not sure how to deal with this then I recommend trying easier composite functions questions before attempting this one.
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    Ok so we know that f(x)=ax+b (equation 1) and ff(x)=9x-28 (equation 2). Using equation one, form an equation for ff(x) and set that equal to equation 2. Then set the x terms equal to eachother.
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    (Original post by james0902)
    Name:  ImageUploadedByStudent Room1472835023.369152.jpg
Views: 150
Size:  97.0 KB

    So this is where I'm at with the working I've put the 100 over to this side but I'm just struggling with solving this! Any help is appreciated (for question 3a)


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    If it helps, f(x) = x^2 and instead of g(x) = 3x - 2, say that y = 3x - 2. This means that g(x) = y

    Now find an expression for f(y). This gives you f(y) = y^2.

    Now, you can substitute y as g(x) because g(x) = y.

    So we get f(g(x)) = (g(x))^2.
    Now substitute g(x) as 3x - 2.

    You get fg(x) = (3x - 2)^2. Set this equal to 100 and solve for x.
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    (Original post by The_Big_E)
    If it helps, f(x) = x^2 and instead of g(x) = 3x - 2, say that y = 3x - 2. This means that g(x) = y

    Now find an expression for f(y). This gives you f(y) = y^2.

    Now, you can substitute y as g(x) because g(x) = y.

    So we get f(g(x)) = (g(x))^2.
    Now substitute g(x) as 3x - 2.

    You get fg(x) = (3x - 2)^2. Set this equal to 100 and solve for x.
    An easier way is fg(a) = 100 -> g(a) = f^-1(100) = +/-10 -> 3x-2 = +/-10 -> 3x = 12 or -8 -> x = 4 or -8/3.
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    fg(x) does not mean multiply f by g(x)
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    (Original post by the bear)
    fg(x) does not mean multiply f by g(x)
    The original poster has already been made aware of that.
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    (Original post by HapaxOromenon3)
    An easier way is fg(a) = 100 -> g(a) = f^-1(100) = +/-10 -> 3x-2 = +/-10 -> 3x = 12 or -8 -> x = 4 or -8/3.
    Ah yes, didn't think of using inverse function. Nice.
 
 
 
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