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@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?

how do i do that?

are u there @electrokid

step 3.1) find the second derivative.

okay

I got 12x^2 + 24x

so i substitute those critical point into the second derivative

Yes.

yes.
\(f''(x)=12x^2+24x\)
step 3.2) plug in x=-3 in this equation and check the sign..

okay

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?

yes

Well, what does that mean, then?

so it will be neither

It will be neither :D

Okay thanks

Already done. ^.^

lol I already did