The function 𝑓(𝑥,𝑦)=𝑥3−3𝑥−2𝑥𝑦−𝑦2−2𝑦+4 has two critical points. Find and classify them

To find the critical points of the function, we first need to find the partial derivatives of the function with respect to x and y.

The partial derivative of f with respect to x is:

f_x = 3x^2 - 3 - 2y

And the partial derivative of f with respect to y is:

f_y = -2x - 2y - 2

Setting these equal to zero gives us the system of equations:

3x^2 - 3 - 2y = 0 -2x - 2y - 2 = 0

Solving this system of equations gives us the critical points.

From the second equation, we can express y in terms of x:

y = -x - 1

Substituting this into the first equation gives:

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

Solving this quadratic equation gives us the x-coordinates of the critical points:

x = [-2 ± sqrt((2)^2 - 43(-1))]/(2*3) x = [-2 ± sqrt(4 + 12)]/6 x = [-2 ± sqrt(16)]/6 x = [-2 ± 4]/6 x = 1/3, -1

Substituting these into the equation y = -x - 1 gives the y-coordinates of the critical points:

y = -(1/3) - 1 = -4/3 y = -(-1) - 1 = -2

So the critical points are (1/3, -4/3) and (-1, -2).

To classify these points, we need to find the second partial derivatives of the function:

f_xx = 6x f_yy = -2 f_xy = -2

The determinant of the Hessian matrix is:

D = f_xxf_yy - (f_xy)^2 D = 6x(-2) - (-2)^2 D = -12x - 4

For the point (1/3, -4/3), D = -12*(1/3) - 4 = -8, which is less than 0, so this point is a saddle point.

For the point (-1, -2), D = -12*(-1) - 4 = 8, which is greater than 0. Since f_xx = 6*(-1) = -6 is less than 0, this point is a local maximum.

So the function has a saddle point at (1/3, -4/3) and a local maximum at (-1, -2).