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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:

    f(x)=1/x²
    g(x)=√(2x-6)

    f(g(x))=1/(√(2x-6))²

    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.

    f(g(x))=1/(2x-6)

    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

    HallsofIvy

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    Staff Emeritus
    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
    [tex]\frac{1}{2x-6}[/tex].

    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

    VietDao29

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    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
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