Question

Find all critical numbers and use the Second Derivative Test to determine all local extrema. y = x^2 - (16)/(x)

Answer 1

@tcarroll010 can u help?

Answer 2

are u there?

Answer 3

ok hold on

Answer 4

y' = 2x + 16/x^2 , y'' = 2 - 32/x^3

Answer 5

i'm trying to click on the equation but its not working

Answer 6

so now i have to set the second derivative = to 0 and solve right?

Answer 7

ok

Answer 8

ok i got x = -1.2599-2.1822 i, x = 2.5198, x = -1.2599+2.1822 i

Answer 9

ok

Answer 10

so what i got is wrong?

Answer 11

ok

Answer 12

so i have correct my derivative right?

Answer 13

to*

Answer 14

ok

Answer 15

idk i dont think i can do this one

Answer 16

i did this one before it asked the same thing but now the problem was this (x^3 - 3x^2 - 9x)
y’ = 3x^2 – 6x – 9
3x^2 – 6x – 9 = 0, x = -1, x = 3
y’’ = 6x – 6
6(-1) – 6 = -12
6(3) – 6 = 12
at -12 it is a maximum and at 12 there is a minimum.

Answer 17

do i do similar to that or?

Answer 18

ok

Answer 19

yea the numerator is equal to zero

Answer 20

ok

Answer 21

yea