Question

g(x,y)=x^4+y^3 has a critical point at (0,0). What sort of critical point it is?

(hint : you can use second derivative test which is : fxx(x0,y0)fyy(x0,y0)-fxy(x0y0)^2 )

Answer 1

do the 2nd derivative test like they suggest, if its positive then it means its increasing

Answer 2

do the gradient of Z=f(x,y)

Answer 3

then do umm divergence of that i think

Answer 4

this is called the laplacian right?

Answer 5

or just do it like they say
fxx = d^2 z/dx^2
fyy=..
fyx=...

Answer 6

plug inot the formula fxx*fyy-fxy^2

Answer 7

the problem is when ı plug into formula it gives me 0 and the formula is useless when the outcom is zero.

Answer 8

thts okay

Answer 9

that's not okay. formula doesnt give the right answer.

Answer 10

it must be a point where the inflection is also happening

Answer 11

so the crit point is of this form

Answer 12

or pointing down

Answer 13

since its either both positive or negative around critical points i think its called an unstable crit point

Answer 14

the answer says they are saddle points?

Answer 15

yeah saddle points

Answer 16

have you seen what the graph looks like on wolfram

Answer 17

http://www.wolframalpha.com/input/?i=z%3Dx%5E4%2By%5E3%2C+x+from+-2+to+2%2C+y+from+-2+to+2

Answer 18

im not sure why they said use 2nd derivative unless there is a thrm saying all unstable crit points in 3D are saddle points

Answer 19

but you can see by dy dz/dx and dz/dy
and findting the crit point and see how it became along the x axis and then along the y axis

Answer 20

and you will find along the x or y axis its increasing and decreasing on the other one