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How to properly find the Domain of fog/gof ?

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    f(x) = x + 1 D=[-9, 9]
    g(x) = x2 D=(1, 5)

    Question to be solved
    Find both g o f and f o g and specify their domains

    The attempt at a solution
    g o f = (x+1)2

    This is where I don't really know what to do, however I'll let the logical part of my brain go a bit silly and try this:
    d= {x e R | 1 < x <= 2} because I'm making the assumption that it has to be true of both domains so while x = -4 would satisfy the domain of f(x) it doesn't g(x). Is that right? If so: how do I show this with math and not just my brain.. I think its inequalities but I've never been good at those, a first step would be much appreciated.

    f o g = x2 + 1
    well x must be >1 but then I need to find whatever value is x2 + 1 < 9
    maybe
    sqrt(8) ? also, as sqrt(8) is clearly < 5 it fits the criteria i think are necessary
    So if I'm correct above it means D{x e R | 1 < x <= sqrt(8)}

    I'm mostly sure I've answered these correctly, but I'm missing the formal way I should be writing this; the last thing I want is marks for correctness and marks lost for not showing my work.
    Thanks to anyone that points me in the right direction.
     
  2. jcsd
  3. Oct 21, 2009 #2

    lanedance

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    i can;t quiet follow what you've done, but you need the range or image of the intermediate function vs the domain of the 2nd function

    so with som sloppy notation
    g(f(x)): x -> f(x) -> g(x)

    now the domain of f is [-9,9], only values of f(x) that fall into (1,5) can be taken by g, so how does this limit the values you can take from [-9,9]

    ie. the domain of g(f(x)) must satisfy both x in [-9,9] and f(x) = x+1 in (1,5)

    then similar for f(g(x))
     
  4. Oct 21, 2009 #3

    Hurkyl

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    When you are composing two partial functions, the "is in the domain" relation can be computed by:
    x is in the domain of gof iff x is in the domain of f and f(x) is in the domain of g​
    Both of the things on the right-hand-side of that iff can be written as inequations, and you know how to solve those, right?
     
  5. Oct 22, 2009 #4
    Okay I'm still not sure with the explanations provided but I'll give it a shot:

    This is the line I'm basing what I've done on:
    "you need the range or image of the intermediate function vs the domain of the 2nd function"
    and I'm thinking about the functions as:

    where:
    f(x) = x + 1 D=[-9, 9]
    g(x) = x2 D=(1, 5)

    and:
    g o f = (f(x))2
    and
    f o g = g(x) + 1

    then:
    g o f = (x+1)2
    D = (1, 5)
    The possible values for f(x) (aka: the range) is [-8, 10] which fails to affect the domain of g

    f o g = x2 + 1
    D = (1, 9]
    The range of g(x) is 1<x<25 limiting the possible Domain of f(x) as it has to be >1

    I hope this is right because it makes sense to me from a logical perspective, if it is I'll chalk this up to another situation of over-thinking.
     
  6. Oct 22, 2009 #5

    lanedance

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    so for the functions, the domain & images are:

    [tex] x \rightarrow f(x) = x+1 [/tex]
    [-9,9] -> [-8,10]

    [tex] x \rightarrow g(x) = x^2[tex]
    (1,5) -> (1,25)

    notice they are both one to one on the given domian which simplifies things

    now for the function gof(x), the image, f([-9,9]) = [-8,10], but g(x) can only take values on (1,5) so we must restrict the domain, otherwise g is undefined, so we look at the f pre-image of the g domain,
    [tex] f^{-1}(x) = x-1 [/tex]
    [tex] f^{-1}((1,5)) = (0,4) [/tex]

    so the total doamin will be given by the intersection of the f pre-image of the g domain & and the f domain
    [tex] D_{gof} = D_f \cap f^{-1}(D_g) = [-9,9] \cap (0,4) [/tex]

    if all this is confusing try drawing 3 horizontal lines above each other.
    - the bottom is x, draw f domain
    - the 2nd is y = (f(x)), draw g domain
    - the 3rd is z = g(f(x))
    now try and draw the approximate mapping action of each function between the lines
     
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