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Homework Help: Domain of f o g

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]f\left( x\right) :=\frac{1}{x-1}[/tex]
    for f(x) Doman (x != 1)
    [tex]g\left( x\right) :=\frac{1}{\sqrt{x+1}}[/tex]
    for g(x), Domain (x >= -1)


    for f(g(x))
    the Domain would be (-1,0) U (0, inf)?

    or do I need to consider other things?
  2. jcsd
  3. Sep 15, 2007 #2
    I was also wondering in composite functions,
    if I find domain after simplifying all the composite function, would that be correct?

    Or I need to find domain restrictions for f(x), and then on g(x)...
  4. Sep 15, 2007 #3


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    The domain is in words: all "x" in the domain of g such as the range of g is in the domain of f. And the answer is indeed (-1,0) reunion (0, +infty).
  5. Sep 15, 2007 #4
    so, do I always need to go that way
    find domain of g(x),

    and any restrictions would be restrictions for f o g

    and then find domain of f(x)
    and any restriction on that would be restriction for the range of g(x)..

    and then domain of f o g = domain of g U (range of g(x) and domain of f(x))..

    or, simply finding domain of f o g from the f of g function - the one I get once I subsitute the g(x) in it?

    Q2/ for inverse functions why we reflect them about y = x func?
  6. Sep 15, 2007 #5
    used vectors to the answer to # 2..

    first dirc vector between (x,y) and (y,x) is (1,-1)

    and thus norm to y =x

    and then distance is always same

    (x-a)^2 + (y-a)^2 = (y-a)^2 + (x-a)^2!!
  7. Sep 15, 2007 #6


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    Yes, but don't simplify the equation once you do the substitution... ie: f(g(x)) = [tex]\frac{1}{\frac{1}{\sqrt{x+1}}-1}[/tex]

    so looking at this function... what's the domain?
  8. Sep 15, 2007 #7


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    looks good to me!
  9. Sep 15, 2007 #8
    I get the same answer, but simplification also gives same domain.

    but I think it would be tedious if they ask for the domain of

  10. Oct 31, 2008 #9
    I also have a relevent/similar question but a bit confusing. We have two functions: f(x) = [tex]\sqrt{1+x}[/tex] and g(x) = [tex]\frac{x}{x-1}[/tex]

    We want to find the domain of f o g.

    D(f) => 1+x2 [tex]\geq[/tex] -1

    D(f) = [-1, [tex]\infty[/tex])

    D(g) = R\{1}

    Now, the domain of f o g is: D(g) such that the g(x) [tex]\in[/tex] D(f).

    => [tex]\frac{x}{x-1} \geq -1[/tex]

    [tex]\frac{2x-1}{x-1} \geq 0[/tex]

    Two points are 1, [tex]\frac{1}{2}[/tex]. Testing between these;

    x<1/2 => +ve
    1/2<x<1 => -ve
    x>1 => +ve

    Therefore the domain of f o g must be [tex](-\infty, \frac{1}{2}) [/tex] U [tex](1, \infty) [/tex]

    The correct answer has to be: D(f o g) = [1/2, 1]. Can anyone explain this for me please?
    Last edited: Oct 31, 2008
  11. Oct 31, 2008 #10


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    Why does the domain "have to be" [1/2, 1]? If x= 3/4, which is in that set, then g(3/4)= (3/4)/(-1/4)= (3/4)(-4)= -3 and f(-3) is not defined. That is essentially the analysis you did. Obviously f(g(x)) is NOT defined at 3/4 and so not defined on [1/2, 1].
  12. Oct 31, 2008 #11
    Yes, I understand that but I checked the answers at the back of the book and it says D(f o g) = [1/2, 1]! That's what makes me confused.
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