Recent content by 0000

mmm... Yeah, maybe it's just what Compuchip says, because if we use the same constant in both intervals it could be thought that a particular primitive must have the same constant in both intervals. Right?

Don't understand why "an indefinite integral is valid only on a interval" Hi I'm using Stewart's Calculus, in the section of indefinite integral, they say: "Recall from Theorem 4.10.1 that the most general antiderivative on a given interval is obtained by adding a constant to a particular...

Thank you for your answers.

Hi, in my physics book (serway) they say "dimensions can be treated as algebraic quantities" but I don't understand this very well. If I sum meters I get meters, if I multiply meters I think I get meters^2 because the area of a rectangle is b.h. But if, for instance, I multiply seconds.seconds...

Ok, thanks, your help has been very useful. And yes, Derive it's a math program, maybe it gives that result because it works with complex numbers and in that case I think that 1/(√(2x-6))² could be equal to 1/2(x-3).

Thanks, I think I see the things a little more clear now, but why when I simplify 1/(√(2x-6))² with Derive I get 1/2(x-3)?

¿Could the domain of a compound function be obtained in the same way that non-compound functions? I think the answer is not, like in this example: f(x)=1/x² g(x)=√(2x-6) f(g(x))=1/(√(2x-6))² Recently Hurkyl explained me that (a^b)^c is not always equal to a^bc, although, I've proved this...

thanks thanks, I think it's my school teacher's fault :P, for teaching me (a^b)^c = a^bc without explaining the constraints of that identity.

In my book and in other places, they give this rule to obtain the domain for a compound function: "the domain of (f o g) (x) is the set of all real numbers x such that g(x) is in the domain of f (x)." Then, if f(x)=x^(1/4) and g(x)=x^2 f(g(x)) = (x^2)^(1/4) f(g(x)) = x^(1/2) And applying...

ok Ok, berkeman X=(Y-2)/3 - 1/(3Y) yx=y((y-2)/3) - 1/(3Y)) yx=(y^2 - 2y)/3 -y/3y) yx=(y^2 - 2y)/3 - 1/3 yx=(y^2 - 2y - 1)/3 y^2 - 2y - 1 - 3yx=0 y^2 - y(2+3x) - 1 = 0 a=1 b=-(2+3x) = (-2 - 3x) c=-1 y=((2+3x)±√(9x^2 + 12x + 8))/2

Thanks, it was 1/(3Y). My fault.

¿How to Find Y? Hi, I have this equation and I need to find Y in terms of X. Could anyone explain me, step by step, how to do it? X=(Y-2)/3 - 1/3Y Sorry if there is any grammar mistakes, english isn't my native language.