Results 1 to 2 of 2

Math Help - find domain

  1. #1
    Member
    Joined
    Sep 2005
    Posts
    136

    find domain

    f(x)=3x-8
    g(x)=3/(x-10)
    find the domain of f+g
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2005
    Posts
    53
    The domain of a function is : all possible x that we can 'plug' in our formula with a result that is a real number (not infinity or square root of a negative number or non-positive logarithm).

    In usual functions, all x work fine. Usually, the problems araise when you have your x in the denominator of a fraction or inside a root (even ones only : square root, forth root...) or when having your x inside a logarithm.

    1) f(x)=3x-8. This function has no problem. I can put any real number x I want and it gives me a value for f(x). Ex. : x=4.5 gives f(4.5) = 3 (4.5) - 8 = 5.5.

    So any real number x works. So the domain is all real numbers. We can write dom(f) = R. Where dom(f) means domain of f and R the set of real numbers.

    2) g(x)=3/(x-10). We see the x in the denominator of a fraction. This tells us that there might be a problem. With what x ? The formula 3/(x-10) craches when x-10 = 0 (we would have 3/0 which is infinity). We know that x-10=0 if and only if x=10. So x=10 is our only problem maker here.

    So all real numbers x except 10 work and give us a result g(x). So the domain of g(x) : dom(g)=(R without 10) =R\{10}. This is just a notation meaning that we remove 10 from the set of real numbers to get the domain of g.
    __________________________________________________ __________

    Now, we add f and g

    (f+g)(x) = 3x - 8 + 3/(x-10). Here we sum so we do not have a new fraction or a square root or a logarithm appearing due to our operation of addition between f and g so no new worries. We just take care of our old worries :

    -) no worries for f
    -) one worry in g : the 10.

    So f+g will not give a resulat for all the x which do not work in f and all those which do not work in g. So in total here we just have 10 which is a problem. So dom(f+g) = (R without 10) = R\{10}.

    If you know intersection of sets I did here intersection of domain of f and domain of g. Because sum does not add more trouble.
    __________________________________________________ _____________

    Here is another example : h(x) = 3/x and m(x) = 3/(x-1).

    dom(h) = R\{0} (we take all real numbers except 0 because h(0)=3/0=infinity) and

    dom(m)=R\{1} for the same reason.

    So h+m = 3/x + 3/(x-1). Both h and m give their problems. So the domain of h+m is R\{0,1} so it is all real numbers except 0 and 1 because (h+m)(0) craches in 3/x and (h+m)(1) craches in 3/(x-1) so craching in one place craches the whole equation.
    Last edited by hemza; November 20th 2005 at 05:59 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the domain
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 10th 2010, 02:46 AM
  2. find the domain of ln(2-5x)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 8th 2009, 04:45 PM
  3. find the domain
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 27th 2008, 01:26 PM
  4. Find Domain
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: June 11th 2007, 04:25 PM
  5. Find the domain
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 11th 2005, 09:32 PM

Search Tags


/mathhelpforum @mathhelpforum