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    Hello.
    So I am doing my C3 Functions homework and I got myself confused at this question:

    f(x)=x^2
    g(x)=2x-1

    Find expression for fg(x).

    Is it fg(x)=(2x-1)^2 or fg(x)=4x^2-1

    Please help!
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    It is your first answer. Remember f o g=f(g(x)) so in your case it's f(2x-1)=(2x-1)^2. f o g is notation for composition of functions if you are wondering. So your second answer would be g o f.
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    (Original post by poorform)
    It is your first answer. Remember f o g=f(g(x)) so in your case it's f(2x-1)=(2x-1)^2. f o g is notation for composition of functions if you are wondering. So your second answer would be g o f.
    Thank you very much!
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    (Original post by saule1116)
    Hello.
    So I am doing my C3 Functions homework and I got myself confused at this question:

    f(x)=x^2
    g(x)=2x-1

    Find expression for fg(x).

    Is it fg(x)=(2x-1)^2 or fg(x)=4x^2-1

    Please help!
    fg(x)=(2x-1)2 is correct but how did you get fg(x)=4x(2-1) ?
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    (Original post by Mehrdad jafari)
    fg(x)=(2x-1)2 is correct but how did you get fg(x)=4x(2-1) ?
    Oops, it was meant to be 4x^(2) - 1... Didn't even realise that until you pointed it out.
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    (Original post by saule1116)
    Oops, it was meant to be 4x^(2) - 1... Didn't even realise that until you pointed it out.
    Oh, I guess that was the only way that could have been since 2-1=1 .
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    (Original post by Mehrdad jafari)
    Oh, I guess that was the only way that could have been since 2-1=1 .
    Haha. ^^
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    (Original post by saule1116)
    Oops, it was meant to be 4x^(2) - 1... Didn't even realise that until you pointed it out.
    You mean 4(x^2)-1 I think.

    Um, usually 4x^2-1 means (4(x^2))-1.

    Technically, 4x^2-1 makes no sense because there are no brackets. We have conventions for where the brackets should go, usually this means we put brackets around the "higher order" operations first.

    Higher order? Well, addition (and its inverse, subtraction) come lowest in the order) because this function grows slowest (some people include the "+1" and "-1" successor and predecessor functions who come lowest of all in the order of functions).

    Then it's multiplication (and its inverse, division) because: axb = a+a+a+a+a...+a {b copies of a}

    Then it's exponentiation (and its inverse, the taking of roots) because: a^b = a x a x a ...x a {b copies of a}

    Actually, this process goes on forever, we define perfectly well "Left tetration" and "Right tetration" as:

    aLb = (...(((((a^a)^a)^a)^a)^a)...)^a {b copies of a}
    and
    aRb= a^...(a^(a^(a^(a^(a^a))))))...) {b copies of a}

    The reason for brackets in the above is because tetration, based on exponentiation, is not "associative":

    It does not obey the rule that (a^b)^c=a^(b^c). For example, what is 2^3^4? It could be (2^3)^4 = 4096 if we sling the brackets to the left. If we sling the brackets to the right, it's 2^(3^4) = about 2.4 billion, billion, billion, billion. A large number.

    So where we put the brackets here is not just a technicality. Perhaps the sheer numerical differences are more convincing when seen in this way. In any case, the result is that the progression from addition, to multiplication, to exponentiation, to left or right tetration, it's not a linear progression, it's actually a tree, of increasingly crazy functions. The closer you position yourself to the root of the tree, the lower order the function. (Technically, the root of the tree is actually not addition/subtraction, but the +1/-1 functions).

    We don't have this problem with 2x3x4 or 2+3+4, and it's here that it starts to get a bit complicated (obviously).

    For example 2L2=2R2=2^2=4.

    But 3L3 = (3^3)^3 = 19683. And 3R3 = 3^(3^3) = about 7.6 billion billion

    Meanwhile 4L4 = (4^4)^4 = 4294967296. And 4R4 = about 1300 million trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion.

    You probably haven't heard of "tetration" because it's so crazy, in fact it exhibites fractal behaviour even in simple cases and is hard to study mathematically.

    The point of this is to illustrate how important brackets are, I suppose. For higher and higher order arithmetic operations, which are of no practical use of course, the actual numbers become less important than where you place the brackets in an expression, in terms of numerical difference.

    Going back to the earlier question, it's very simple: put a bracket around the highest order operation first. So start with 4x^2-1. This turns into:

    4(x^2)-1.

    Now put brackets around the next highest order operation. So the above becomes:

    (4(x^2))-1.

    This is basically how bidmas or whatever it's called works.
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    (Original post by poorform)
    It is your first answer. Remember f o g=f(g(x)) so in your case it's f(2x-1)=(2x-1)^2. f o g is notation for composition of functions if you are wondering. So your second answer would be g o f.
    No it wouldn't

    g(f(x))=g(x^2)=2x^2-1
 
 
 
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