Recent content by Velo

Krylov asked me to give an example of one $b∈{\Bbb{R}}^{3}∈$ for which $(A|b)$ is not consistent... With A equal to the second matrix I assumed... So if the last position of the b vector is non zero (1 in my example), the equation would be something like $0{x}_{1}+0{x}_{2}+0{x}_{3}=1$, or $0 =...

Something along the lines of ${[0, 0, 1]}^{t}$? Or really any vector with a third position different from zero...

Yes, I apologize. English is not my native language either, and most of the mathematical concepts I know are in Portuguese, so I have a rough time translating them sometimes :') I think my confusion comes mostly from when a matrix equation ends up having something like $0=0$ though...

But why is the column space's dimension $r(A)$? Wouldn't that be assuming that the system is linearly independent? Is that what they meant when they said that $Ax=b$ is possible/solvable? 'Cause the way I interpreted it, I thought that all (1) was saying is that the system was consistent... Did...

So, my linear algebra book, if you can call it that, says the following: $Ax=b$ is a system of linear equations with $m$ equations and $n$ variables. ${v}_{1}, {v}_{2}, ..., {v}_{n}$ are the vectors in the columns of $A$. The following are equivalent: (1) The system $Ax=b$ is possible for...

Classic :')

OOOhhhhh! I see! So, $f'(-2) = -2 \implies y = -2x + b$, From the original f(x) we also know that $f(-2) = 5$ so we can just replace that int he equations and get b! $y = -2x + b \implies 5 = (-2) * (-2) + b \Leftrightarrow b = 1$ And so the final equation is: $y = -2x + 1$ So if we wanted to...

I'm not sure I'm following... I know a Taylor expansion should look something like this when we're trying to get a function similar to another function on $x = -2$: $f(-2) + f'(-2)(x+2) + \frac{f''(-2)}{2}(x+2)^2 + o(x+2)^2$ I also get that this new function will be very close to the original...

So, the information they give me is the following: $(1) f \in {C}^{3}({\rm I\!R})$ $(2) f(x) = 5 -2(x+2) - (x+2)^2 + (x+2)^3 + R3(x+2)$ $(3) \lim_{{x}\to{-2}} \frac{R3(x+2)}{(x+2)^3}=0$ And they ask me for the equation of the tangent line... Which would be simple if that R3 wasn't there...

Thanks a lot :3 Was really struggling with this for some reason, even though it was actually pretty simple >..<

Oh, I think I got it now... So in the equation y = mx + b, b is the y when x = 0 in that equation, correct? :o And then, since our starting point in the tangent line's equation doesn't actually have to bex = 0, we move around that point instead?

So, I can't wrap around my head of why the Equation of the Tangent Line is: y = f(a) + f'(a)(x - a) I get it that it's the equation of a line, and so it should be something like y = mx + b. I also understand why f(a) = b (since it's a point in that line) and why f'(a) = m (since it's the slope)...

Thanks for the quick reply! Sorry I created the post in the wrong forum btw, I wasn't sure where limits would go to :S Also, hm.. Should we always use the substitution method to make the variable tend to +\infty instead of -\infty? Or is there something about that particular limit that makes...

So, I'm still struggling with limits a bit.. Today, I've tried solving two different exercises which look pretty much the same. I could solve the first one relatively easily: \lim_{{x}\to{+\infty}}\frac{\sqrt{4x^{2}-1}-x}{x-3} I applied the usual steps and arrived to the expression...

Ohhh, I get it now.. I read the book wrong too :') The solution had \log_e(\frac{2}{3}) and not \log(\frac{2}{3}).. I spent so much time wondering where that log had come from :') I tried redoing that exercise and the next in my notebook and I'm doing alright now x3 Thank you very much! :D