Stationary points of a function of two variables.?

f(x,y) = 1-y^3-3yx^2-3y^2-3x^2

∂f/∂x=-6xy-6x =0 ----(1)

y=-1

∂f/∂y=-3y^2-3x^2-6y=0 ---(2)

-3(1)-3x^2+6 =0

-3x^2=-3

x^2=1

x=+/- 1

Stationary points x=1, -1 & y=1

(∂^2f/∂x^2) = -6y

(∂^2f/∂y^2) =-6y-6

(∂^2f/∂x∂y) = -6x

Substitute x=1, y=-1

Find D(x,y)=(∂^2f/∂x^2) (∂^2f/∂y^2) - (∂^2f/∂x∂y)^2

D(x,y) =6(-12)-36 < 0

If D(a,b) > 0 and ∂^2f/∂x^2 > 0, then f(x,y) has a

maximum at (a,b)

If D(a,b) > 0 and ∂^2f/∂x^2 < 0, then f(x,y) has a

minimum at (a,b)

f has a minimum at (1,-1)

You can similarly prove that f has a maximum at (-1,-1)

If D(a,b) < 0 and , then f(x,y) has neither a maximum nor a

minimum at (a,b)

If D(a,b) = 0 and , then no conclusion can be drawn from this

limit

regrettably i don't have a application on my workstation able to plotting 3-d surfaces, so that's a query of taking the bull via the horns and working it out the complicated way. For f(x, y) = [3/(2.x) - x/2 - y]² + x² + y² ?f/?x = 2.[- 3/(2.x²) - a million/2].[3/(2.x) - x/2 - y] + 2.x . . . = - [3/x² + a million].[3/(2.x) - x/2 - y] + 2.x ....(a million) ?f/?y = - 2.[3/(2.x) - x/2 - y] + 2.y = - (3/x - x) + 4.y ....(2) putting eq.(2) equivalent to 0 supplies y = 3/(4.x) - x/4. ....(3) Substituting this into eq.(a million) and putting this to 0 yields ?f/?x = 0 = - [3/x² + a million].[3/(4.x) - x/4] + 2.x . . . . . . = [- 9/(4.x³) + 3/(4.x) - 3/(4.x) + x/4] + 2.x = - 9/(4.x³) + 9.x/4 which reduces to x^4 = a million or x = ±a million. Substituting those values into eq.(3) supply for x = a million, y = ½ and for x = -a million, y = -½, and those are the only 2 table certain factors present day. to be sure their nature, we desire the 2nd partial differentials. Multiplying by and rearranging eq.(a million) supplies ?f/?x = - [3/x² + a million].[3/(2.x) - x/2] + 2.x + y.(3/x² + a million) . . . .= - 9/(2.x³) + 3/(2.x) - 3/(2.x) + x/2 + 2.x + y.(3/x² + a million) . . . .= - 9/(2.x³) + 5.x/2 + y.(3/x² + a million) Differentiating wrt x ?²f/?x² = 9/(2.x^4) + 5/2 - 6.y/x³ ....(4) Differentiating eq.(2) wrt y supplies the easy result ?²f/?y² = 4. Substituting the coordinates of the two table certain factors in eq(4) produces an identical result ?²f/?x² = 13 for the two, so we end that for the reason that all the 2nd differentials are effective, the two factors are interior of sight minima.

not the answer but some help

f(x, y)= 0 is equivalent to y = f(x).

1- y^3 -3yx^2 - 3 y^2 - 3 x^2 = 0.

(1-y^3 -3y^2) - 3x^2( 1+ y) = 0.

x^ 2 = (1-y^3 - 3 y^2)/3(1+y).

use the property of calculus dy/ dx = 0. to locate stationary points, local max. and min.

hope it helps.

i know that stationary points are when the gradient equals zero so if you know how to find that then maybe youll find the answer. when we learnt that we used a graphics calculator where we could sketch the function to make it easier to find answers