The function 𝑓(𝑥,𝑦)=𝑥3−3𝑥−2𝑥𝑦−𝑦2−2𝑦+4 has two critical points. Find and classify them. the left-most stationary point (the one with the lower value of x) is located at x = and y = and is a point. • the right-most stationary point (the one with the higher value of x) is located at x = and y = and is a point

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, and then set them equal to zero.

The partial derivative of f with respect to x is:

f_x = 3x^2 - 3 - 2y

Setting this equal to zero gives us:

3x^2 - 3 - 2y = 0

The partial derivative of f with respect to y is:

f_y = -2x - 2y - 2

Setting this equal to zero gives us:

-2x - 2y - 2 = 0

Solving these two equations simultaneously will give us the critical points of the function.

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

y = -x - 1

Substituting this into the first equation gives us:

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

Solving this quadratic equation gives us:

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 x = 1/3 into y = -x - 1 gives y = -1/3 - 1 = -4/3 Substituting x = -1 into y = -x - 1 gives y = 1 - 1 = 0

So the critical points are (1/3, -4/3) and (-1, 0).

To classify these points, we need to find the second partial derivatives of the function and evaluate them at the critical points.

The second partial derivative of f with respect to x is:

f_xx = 6x

The second partial derivative of f with respect to y is:

f_yy = -2

The mixed partial derivative of f with respect to x and y is:

f_xy = -2

The determinant of the Hessian matrix at a point (x, y) is given by:

D = f_xx * f_yy - (f_xy)^2

Evaluating D at the point (1/3, -4/3) gives:

D = 6*(1/3) * -2 - (-2)^2 D = -4 - 4 D = -8

Since D < 0, the point (1/3, -4/3) is a saddle point.

Evaluating D at the point (-1, 0) gives:

D = 6*(-1) * -2 - (-2)^2 D = 12 - 4 D = 8

Since D > 0 and f_xx < 0, the point (-1, 0) is a local maximum.

So, the left-most stationary point (the one with the lower value of x) is located at x = -1 and y = 0 and is a local maximum. The right-most stationary point (the one with the higher value of x) is located at x = 1/3 and y = -4/3 and is a saddle point.