Question

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

Answer 1

@electrokid can u help?

Answer 2

using our previous discussions, you can start the work and I guide.

Answer 3

step 1) first derivative
\(y'=?\)

Answer 4

Sorry i replied late i had internet issues. but the first derivative is 4x^2(x + 3)

Answer 5

good.
step 2) solve for "x":
\(4x^2(x+3)=0\)

Answer 6

okay i got x = -3 and x = 0

Answer 7

are you there?

Answer 8

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)

Answer 9

how do i do that?

Answer 10

are u there @electrokid

Answer 11

step 3.1) find the second derivative.

Answer 12

okay

Answer 13

I got 12x^2 + 24x

Answer 14

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.

Answer 15

so i substitute those critical point into the second derivative

Answer 16

Yes.

Answer 17

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

Answer 18

okay

Answer 19

i got 36

Answer 20

and when i plugged in 0 i got 0

Answer 21

Okay, so, f''(-3) = 36
This is positive, therefore...?

Answer 22

so it is a minimum?

Answer 23

a local minimum ;)
What about f''(0) ?

Answer 24

I got 0

Answer 25

It's neither positive nor negative, right?

Answer 26

yes

Answer 27

Well, what does that mean, then?

Answer 28

so it will be neither

Answer 29

It will be neither :D

Answer 30

Okay thanks

Answer 31

@PeterPan and @onegirl both did a great job and effort and both deserve a medal :)
so, @onegirl give one to P-Pan
☮

Answer 32

Already done. ^.^

Answer 33

lol I already did