1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Another question about the domain in a compound function

  1. Jun 24, 2007 #1
    ¿Could the domain of a compound function be obtained in the same way that non-compound functions?

    I think the answer is not, like in this example:



    Recently Hurkyl explained me that (a^b)^c is not always equal to a^bc, although, I've proved this step with Derive and it seems to be fine.


    Back to the matter, the domain of the function H(x)=1/(2x-6) is R - {3}

    But the domain of f(g(x))=1/(2x-6) is (3,∞)

    So, Am I right in the fact that a compound function can have a different domain than the "same function" which isn't the product of a composition of functions?

    As always, excuse me if my english isn't very clear.
    Last edited: Jun 25, 2007
  2. jcsd
  3. Jun 25, 2007 #2


    User Avatar
    Science Advisor

    Note that a function has two parts- a domain (possible values of x) and a rule connecting each x with a unique a. That is "f(x)= x2 for x> 0" is not the same function as "f(x)= x2". In the formula
    [tex]f(g(x))= \frac{1}{(\sqrt{2x-6})^2}[/tex]
    since the domain of g is[itex] x\ge 3[/itex], the square root cannot be taken for x< 3 and so the domain is [itex]x\ge 3[/itex].

    [tex]f(g(x))= \frac{1}{(\sqrt{2x-6})^2}[/tex]
    is NOT the same function as

    This is the reason Calculus texts refer to the "deleted" neighborhood of a when taking limits at a. In order to take [itex]\lim_{x\rightarrow 2} (x^2-4)/(x-2)[/itex] you must note that [itex](x^2-4)/(x-2)[/itex] for all x except x= 2 and use the theorem "If f(x)= g(x) in some deleted neighborhood of a, then they have the same limit at x= a".
  4. Jun 25, 2007 #3
    Thanks, I think I see the things a little more clear now, but why when I simplify
    1/(√(2x-6))² with Derive I get 1/2(x-3)?
  5. Jun 25, 2007 #4


    User Avatar
    Homework Helper

    Err... What's Derive? Some computer programme, right? Btw, just don't rely on them so much. :)

    Ok, here it goes.

    In fact, the statement:
    [tex]\left( \sqrt{x} \right) ^ 2 = x[/tex] is incorrect.

    When saying f(x) = g(x), it must be true that:
    1. f(x), and g(x) have the same domain.
    2. They (f(x), and g(x)) return the same value for every x in their domain.

    In this case:
    [tex]\left( \sqrt{x} \right) ^ 2 = x[/tex] violates the first requirement.

    The LHS (left-hand side) function is only defined for x >= 0, while the RHS (right-hand side) is defined for all x in the reals.

    But, however, if you restrict the domain for the RHS function to be [tex][0 ; \ \infty [[/tex], then they can be said to be equal, i.e: [tex]\left( \sqrt{x} \right) ^ 2 = x[/tex]


    So, say, we let [tex]f(x) = \sqrt (x) ; \quad g(x) = x ^ 2 ; \quad h(x) = x[/tex], then:
    f(g(x)) does not have the same domain as h(x).

    Can you get it? :)
    Last edited: Jun 25, 2007
  6. Jun 25, 2007 #5
    Ok, thanks, your help has been very useful. And yes, Derive it's a math program, maybe it gives that result because it works with complex numbers and in that case I think that 1/(√(2x-6))² could be equal to 1/2(x-3).
  7. Jun 26, 2007 #6
    VietDao made the point that "Derive" is correct iff x ≠ 3. I'd also like to point out that if [tex](\sqrt{x})^2[/tex] is a complex valued function, then [tex](\sqrt{x})^2[/tex] = x for all x £ C.

    In general, if f(z) is a complex valued function and f(z) = g(z) is a relation, then f(z) = g(z) for all z £ C.

    If f(x) is a real valued function and f(x) = g(x) is a relation, then f(x) = g(x) for x £ D, where D is the real domain.
    Last edited: Jun 26, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook