1. ## Critical point question

hi everyone

having problems finding critical points for this two equation to fined the maximum,minimum or saddle points.

a) $x^4+y^4-4xy+1$
$f_x=4x^3-4y$, $f_y=4y^3-4x$

b) f(x,y)= $x^4-2x^2y+2y$
$f_x=4x^3-4xy$, $f_y=2-2x^2$

need help to find the critical points for this two equations,really hope someone can help me. really appreciate all your help & support.

2. How far have you come in your attempts? Where do critical points for a given function occur?

a) $x^3-y=0$, $y^3-x=0$

$x^3=y$
$y^3=x$
im stuck here, unable to find the critical point
b) $4x^3-4xy=0$
$4x^3=4xy$
$x^2=y$

$2-2x^2=0$
$2=2x^2$
x=1 & -1

critical point, = (1,1) & (-1,1) ..is this right???

realyl hope someone can help, appreciate all help & support.

4. a: Remember that the two derivatives form a system of equations:
$x^3-y = 0$
$y^3-x = 0$

imply that

$x^9=x$ or $x(x^8-1) = 0$ (see why?)

What values for critical points does this give you?

b: Seems right to me, so far, however, you are being a little hasty. It's quite right that
$4x^3-4xy = 0$ implies that
$x^2 = y$
but it also implies that
$x(x^2-y) = 0$. See what you missed?

a)
$(x^3)^3-x=0$
$x(x^8-1)$
x= -1,0 & 1
critical points=(-1,-1)(0,0) & (1,1)
b)you are right, but is my final answer for critical point correct?
critical point=(1,1) & (-1,1) .. i think it has to be correct..not sure,a bit confused though

is this correct?

thank you again for all help & support.