1. ## Find maxima, mimima and saddle points (fraction)

Find maxima, mimima and saddle points for each of the following func-
tions:
a) f(x,y)=6x^2 - 2x^3 + 3y^2 + 6xy
b) f(x,y)=(x^2 + y^2 - 1)^-1

Firstly, is my answer for part a) correct:
df/dx= 12x - 6x^2 + 6y
df/dy= 6y + 6x

found x=0 and x=1 then y=0 and y=-1
so, the critical points are (0,0) and (1.-1)

doing the 'second derivative test' found (0,0) to be a local minimum and (1,-1) to be a saddle point.

Secondly, i am at a total loss of what to do for part b) as its a fraction and i cant seem to use partial fractions to seperate the variables!!?

Thank you x

2. Originally Posted by cheeseontoast
Find maxima, mimima and saddle points for each of the following func-

tions:

b) f(x,y)=(x^2 + y^2 - 1)^-1

Secondly, i am at a total loss of what to do for part b) as its a fraction and i cant seem to use partial fractions to seperate the variables!!?

Thank you x

If I am not wrong, it is:
df/dx = -(2x)/(x^2+y^2-1)^2
df/dy = -(2y)/(x^2+y^2-1)^2 ,isnt it?

3. Originally Posted by Kiki
If I am not wrong, it is:
df/dx = -(2x)/(x^2+y^2-1)^2
df/dy = -(2y)/(x^2+y^2-1)^2 ,isnt it?
ok thanks. and what about the rest of the question (finding critical points and maxima/minima)?
how do i find d2f/dx2 and d2f/dy2?
Thanks