Domain of a function, the 3 common cases to be careful! - Videos


How to find the domain of a function?
Here I go over the 3 situations (with 6 examples) that you must know!
part 2:

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  1. Let f(x) = (x^2-1)/(x^2+3x+2)

    Bottom = x^2+3x+2 = (x+1)(x+2)

    So x ≠ -1 and x ≠ -2

    But f(-1) = ((-1)^2-1)/((-1)^2-3+2) = 0/0

    Using L'Hospitals rule we get g(x) = 2x/(2x+3)

    g(-1) = -2/(-2+3) = -2

    So Domain is ( x ≠ -1 and x ≠ -2 ) or only x ≠ -2?

  2. I've learned following notations:

    If the underlying sets of the examples are R:

    Ex.1: D=R{-2, -1}
    Ex.2: D=R
    Ex.3: D={x∈R|x≤5}
    Ex.4: D={x∈R|x<5}

    By the way: In German this D stands for "Definitionsbereich"

  3. I think for f(x)=x^x the domain would be:

    D=R⁺∪(Q⁻{x∈Q|x=-p/(2q) where p∈N*, q∈N*, gcd(p, 2q)=1, })

    With this domain zero (since 0^0 is undetermined), negative fractions with even denominators and negative irrational numbers like -√2 or π are excluded.

    So x=-1/3 is alright: (-1/3)^(-1/3) = 1/((-1/3)^(1/3)) = 1/∛(-1/3) = -1,4422495703074083823216383107801

  4. here's an unusual, yet very useful function:

    BPRP : Mathematics -> YouTube
    BPRP ( Subject ) = Video on Subject
    Sadly, the BPRP function is neither injective, nor surjective, but it has some intriguing properties, such as:
    – it makes Subject more accessible for all Subject from Mathematics
    – it has strong ties with the Mathematics-defined branch of the Humor function
    – sgn(BPRP(Subject)) = Positive for all Subject from Mathematics
    – the integral of BPRP, CommunityOfBPRP, is is defined for all Subject in Mathematics and is positive everywhere
    – the feedback property: BPRP and CommunityOfBPRP are commutable in composition and produce the Subject as a result for all Subject in Mathematics
    – BPRP ( BPRP ( Subject) ) =CommunityOfSubject( Subject )
    – it has a strong link to the PeyamShow function (which has the same domain and codomain), that is, the derivative of BPRP is the integral of PeyamShow

    Also, the value of the function as the subject gets more and more abstract seems to be asymptotic to some strange function in some brown/blue space , the limit as abstractness grows without bound tends to 3.

  5. @ 9:40 assume the ≧ inequality symbol shouldn't change upon multiplying or dividing by negative numbers, let x=1
    if x = 1
    Proof by contradiction shows that the inequality signs must be changed to their opposite upon a negative multiplication or division. ≦ ≧ < >
    The real question is what do inequalities with complex numbers look like!!!!?????? 我的天啊????!!!! #YAY

  6. Hey blackpenredpen. I love your videos and always look forward to when you upload. I couple of days ago I found an algebra problem that I would love to see you solve in a video. It is:

    What is the value of ‘y’ so that y times x^1/sqrt(x) = x

    If you could do this in a video I would really appreciate it!


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