Find all critical numbers and use the First Derivative Test to classify each as the location of a location maximum, local minimum, or neither. y = x^4 + 4x^3 - 2

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Find all critical numbers and use the First Derivative Test to classify each as the location of a location maximum, local minimum, or neither. y = x^4 + 4x^3 - 2

Mathematics
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@electrokid can u help?
using our previous discussions, you can start the work and I guide.
step 1) first derivative \(y'=?\)

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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.
okay
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.
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
@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

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