At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
@electrokid can u help?
using our previous discussions, you can start the work and I guide.
step 1) first derivative \(y'=?\)
Sorry i replied late i had internet issues. but the first derivative is 4x^2(x + 3)
good. step 2) solve for "x": \(4x^2(x+3)=0\)
okay i got x = -3 and x = 0
are you there?
good. those are the critical points. step 3) classify as maxima, minima or neither step 3.1) find second derivative of "f" -> find f''(x)
how do i do that?
are u there @electrokid
step 3.1) find the second derivative.
I got 12x^2 + 24x
Remember the critical points that you got? If the second derivative is positive when evaluated at one of those critical points, that critical point would be a local minimum. On the flip-side, if the second derivative is negative when evaluated at a critical point, that point would be a local maximum.
so i substitute those critical point into the second derivative
yes. \(f''(x)=12x^2+24x\) step 3.2) plug in x=-3 in this equation and check the sign..
i got 36
and when i plugged in 0 i got 0
Okay, so, f''(-3) = 36 This is positive, therefore...?
so it is a minimum?
a local minimum ;) What about f''(0) ?
I got 0
It's neither positive nor negative, right?
Well, what does that mean, then?
so it will be neither
It will be neither :D
@PeterPan and @onegirl both did a great job and effort and both deserve a medal :) so, @onegirl give one to P-Pan ☮
Already done. ^.^
lol I already did