Math Is Fun Forum

yes

Well, I think I could show you an example of what I was talking about. I read it somewhere :

We have :

a^2+2ab+b^2

Now we know that the first term of the root will be "a"

so...

a^2+2ab+b^2(a

"a" squared cancels the other a^2

2ab+b^2(a

Now, we know that 2ab+b^2=(2a+b)b and so if we can figure out a divisor, we will also know the root. We also know the first term of the divisor because it is always the double of the firm term of the root. (Look at the square root algorithm :http://www.basic-mathematics.com/square-root-algorithm.html)

2ab+b^2(a+b           (we find, by luck I guess, that we have "b" for the second term.)
____________
2a+b)2ab+b^2
         2ab+b^2
______________
               0

Anyway, i was searching for something like what I described.

Ok, byt is there a sort of long division method we could use ???

Nah, I don' even know what is a CAS. But I would want to do it by hand....

Ok, well if I gave you : (a+b-c)(a+b-c) or (a+b-c)(a+b-c)(a+b-c), each in developped form and I asked you what were the square root of the first and the cube root of the second, you would have : (a+b-c) and (a+b-c).

I want to extract the root in developped form!

Ok, bu they did an error, we should read y --> -2^(-) and not y --> 2^(-)

The same thing goes for : y --> -2^(+) and not y --> 2^(+)

Exactly!Anyway, I need to go now. Thanks (for the millionth time) for the dedication!

Ok, thank you again for all. For some reason, my head wasn't here today... Do you want to know why I kept bugging ? I assumed, against my will, that limit and function value (the point c) were the same thing.... I don't why I did this.... Anyway...

Oh, I wouldn't want to annoy you, but I have another question.

If we don't consider my graph as two separate functions(So, we don't have a function anymore, only a relation.So it doesn't pass the vertical line test), and we still want to evaluate the limit at the same point, it would still be acceptable, even if its not a function?

Ok, then I understod the fact. Now, concerning my function questiom, if we take my graph, we see that at the point x=-3 we have two y's. The black and red line are both functions. But with the vertical line test, we see that this isn't a function. So, my question is, this "fact" about the inequalities doesn't necessarily need the whole (non piece wise function) to be a function.

Ok, but my question is, does this "sum" of function need to be a function itself ? We can have piecewise functions and non piecewise function. In other words, can we evaluate limits of non piece wise function (It doesn't satisfy the vertical line test.)