Question

Compute the discriminant D(x,y)D(x,y) of the function. f(x,y)=x3+y4−6x−2y2+3 f(x,y)=x3+y4−6x−2y2+3 (Express numbers in exact form. Use symbolic notation and fractions where needed.) D(x,y)=D(x,y)= Which of these points are saddle points? (2⎯⎯√,0)(2,0) (2⎯⎯√,1)(2,1) (−2⎯⎯√,1)(−2,1) (2⎯⎯√,−1)(2,−1) (−2⎯⎯√,−1)(−2,−1) (−2⎯⎯√,0)(−2,0) Which of these points are local minima?

Answer 1

Answer:

Point                Critical point

Q (2,0)                 local minimum

R (-2,1)                 Saddle

S (2,-1)                  local maximum

T ( -2,-1)                Saddle

O ( -2,0)                Saddle

Step-by-step explanation: INCOMPLETE ANSWER INFORMATION ABOUT THE POINTS ARE RARE

f(x,y) = x³ +y⁴ - 6x -2y² +3

df/dx = f´(x) = 3x² -6x

df/dxdx = f´´(xx) = 6x

df/dy = f´(y) = -4y

df/dydy = 4

df/dydx = df/dxdy = 0

df/dydy = f´´(yy)

D = [ df/dxdx *df/dydy] - [df/dydx]²

D = (6x)*4 - 0

D = 6*2*4     D > 0 and the second derivative on x is  6*2 = 12

so  D > 0 and df/dxdx >0  there is a local minimum in P

Q(2,1)

D = (6*2)*4  D>0  and second derivative on x is 6*2

The same condition there is a minimum in Q

R ( -2,1)

D = 6*(-2)*4 = -48  D< 0 there is a saddle point in R

S (2,-1)

D = 6*2*4 = 48  D > 0 and  df/dxdx = 6*-1  = -6

There is a maximum in S

T ( -2,-1)

D = 6*(-2)*(4) = -48  D<0 there is a saddle point in T

O ( -2,0)

D = 6*(-2)*4 = -48 D<0  there is a saddle point in O