Question

find the critical numbers of 4y^3 +2y + 6

Answer 1

In the second term is it y or y^2

Answer 2

If it is y itself then no critical pints since first derivative is never zero in the set of real numbers

Answer 3

original function is (-y^2)^2 +y^2+6y+9
f '(y) = 4y^3+2y+6
f ''(y) = 12y^2+2

Answer 4

Ohhh u have to mention than first dude I thought that it is the main function

Answer 5

One of the critical point is -1

Answer 6

i figured that after i read your response.. my bad..

Answer 7

Its k

Answer 8

im just having issues working out the first derivative to find my critical numbers..

Answer 9

And the other two are imaginary

Answer 10

differentiate the equation first
12y^2+2
then equate it to 0
12y^2+2=0
y^2= -1/6
aww man!!!!!!!!!!!!!!!!!
its critical points doesn't exits

Answer 11

DHASHINI once read our replies tooo u will get the info about the question

Answer 12

ok.. my math could be wrong then. perhaps i should had started with my problem.
find the point on the parabola x + y^2 =0 that is closest to (0,-3)

Answer 13

Dude what you did was absolutely correct just go on we have only one critical point this may be the shortest too wait a second i will let u know

Answer 14

Ya it is the point of minimum because the value of y=-1 gives positive for second derivative and substitute this value of y dude

Answer 15

Point is (-1,-1) And surely it lies on the parabola also dude

Answer 16

and the shortest distance is 5 units

Answer 17

ok.. thanks.. i was certain the word point was a typo.. i was counting on the parabola to have 2 point(s)

Answer 18

no probs