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Math Help - Accounting for restricted x-values in fractional (linear/linear) derivatives

  1. #1
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    Accounting for restricted x-values in fractional (linear/linear) derivatives

    Hello everyone,

    I was working on a problem (solution shown):

    and checking my work with wolfram alpha to find that there is a "restricted x-value" constant that is added in to the end of the integral, in this case 4/9. I don't know how this value was calculated, but Wolfram has the steps it took here: http://www.wolframalpha.com/input/po...e46hi91b8&s=57

    Could someone possibly explain the use of this constant and how it was calculated, since it can be eliminated anyway if C=-4/9?

    Also, are there other cases when this need be calculated? Could you point me to some places where I can get some practice?

    Help is greatly appreciated!
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  2. #2
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    Looking at the 3rd term

    \frac{1}{9}(6x+5\log(3x+2)+4)+C=\frac{6x}{9}+\frac  {5\log(3x+2)}{9}+\frac{4}{9}+C

    The restricted value here refers to the \log part of the function as \log(x), x>0
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Looking at the 3rd term

    \frac{1}{9}(6x+5\log(3x+2)+4)+C=\frac{6x}{9}+\frac  {5\log(3x+2)}{9}+\frac{4}{9}+C

    The restricted value here refers to the \log part of the function as \log(x), x>0
    Alright, so one plugs in -2/3 so that the log function goes to 1, and the entire function (pre-restriction term) goes to 4/9, which is then added in as the restriction term.

    1. Why is 0 chosen to be plugged into the equation?
    2. Whats the purpose of the restriction term if it can just be eliminated if C=-4/9?

    Thanks!
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  4. #4
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    Quote Originally Posted by progressive View Post

    1. Why is 0 chosen to be plugged into the equation?

    Thanks!
    Where was 0 plugged in? I do not follow this question.


    Quote Originally Posted by progressive View Post

    2. Whats the purpose of the restriction term if it can just be eliminated if C=-4/9?

    Thanks!
    How is c = -4/9? Where you given some initial conditions?
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  5. #5
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    Where was 0 plugged in? I do not follow this question.
    I meant that a value of x was plugged into the function so that it would satisfy log (0), which would effectively eliminate the term with the log in it, and when that same x value is plugged into the other term (I believe x=-2/3 satisfies this), then one ends up with 4/9. At least thats how I think you get 4/9. How do you really get it? Thats what I asked originally.

    How is c = -4/9? Where you given some initial conditions?
    I don't see the point of adding a constant to account for restricted x-values when the constant can be theoretically eliminated if C is the opposite of that constant. C does not have to be -4/9, but if it were then it would just cancel out the restrictive term, so I'm asking what the point of the restrictive term is. I don't really understand what it accounts for either. Could you please explain?
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  6. #6
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    Quote Originally Posted by progressive View Post
    so that it would satisfy log (0),
    You can try to satisfy log(0) if you like and in your case yep, x = -2/3. But what does log(0) give you? Put it in your calculator. Have a look.

    I am saying (read above) \log(x), x>0 not \log(x), x=0

    In your case \log(3x+2), x>\frac{-2}{3} not \log(3x+2), x=\frac{-2}{3}

    This part x>\frac{-2}{3} is the restriction.

    It has nothing to do with the C=\frac{4}{9}


    Quote Originally Posted by progressive View Post


    I don't see the point of adding a constant to account for restricted x-values when the constant can be theoretically eliminated if C is the opposite of that constant. C does not have to be -4/9, but if it were then it would just cancel out the restrictive term, so I'm asking what the point of the restrictive term is. I don't really understand what it accounts for either. Could you please explain?
    Neither do I. I can't see how these ideas are related at all. C is is your constant of integration. It must be included. Do you know why?

    If you like you can say.

    \frac{1}{9}(6x+5\log(3x+2)+4)+C=\frac{6x}{9}+\frac  {5\log(3x+2)}{9}+\frac{4}{9}+C =\frac{6x}{9}+\frac{5\log(3x+2)}{9}+C


    Hope this helps some. I understand it can be tricky!
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