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

• Sep 5th 2010, 10:19 PM
progressive
Accounting for restricted x-values in fractional (linear/linear) derivatives
Hello everyone,

I was working on a problem (solution shown):
http://www4d.wolframalpha.com/Calcul...=57&w=344&h=37
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!
• Sep 5th 2010, 10:29 PM
pickslides
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 $\log(x), x>0$
• Sep 6th 2010, 06:51 AM
progressive
Quote:

Originally Posted by pickslides
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!
• Sep 6th 2010, 02:24 PM
pickslides
Quote:

Originally Posted by progressive

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

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?
• Sep 6th 2010, 02:48 PM
progressive
Quote:

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.

Quote:

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?
• Sep 6th 2010, 03:55 PM
pickslides
Quote:

Originally Posted by progressive
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

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!