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

1. ## 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!

2. 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 $\log(x), x>0$

3. 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!

4. 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.

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?

5. 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?

6. 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 $C=\frac{4}{9}$

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!