To find the critical points of the function f(x, y) = x³ - y² + 9xy + 5, 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 = 3x² + 9y. The partial derivative of f with respect to y is f_y = -2y + 9x. Setting these equal to zero gives us the system of equations: 3x² + 9y = 0, -2y + 9x = 0. Solving this system of equations gives us the critical points. From the second equation, we can express y in terms of x: y = 4.5x. Substituting this into the first equation gives us 3x² + 9*4.5x = 0, which simplifies to 3x² + 40.5x = 0. This can be factored to 3x(x + 13.5) = 0, giving us the solutions x = 0 and x = -13.5. Substituting these values back into y = 4.5x gives us the corresponding y-values: y = 0 and y = -60.75. So, the critical points are (0, 0) and (-13.5, -60.75). To determine the nature of these points, we need to find the second partial derivatives of f: f_xx = 6x, f_yy = -2, f_xy = f_yx = 9. The determinant of the Hessian matrix, D = f_xx*f_yy - (f_xy)², will tell us the nature of the points. If D 0 and f_xx 0, the point is a local minimum. If D 0 and f_xx

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To find the critical points of the function f(x, y) = x³ - y² + 9xy + 5, 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 = 3x² + 9y. The partial derivative of f with respect to y is f_y = -2y + 9x. Setting these equal to zero gives us the system of equations: 3x² + 9y = 0, -2y + 9x = 0. Solving this system of equations gives us the critical points. From the second equation, we can express y in terms of x: y = 4.5x. Substituting this into the first equation gives us 3x² + 9*4.5x = 0, which simplifies to 3x² + 40.5x = 0. This can be factored to 3x(x + 13.5) = 0, giving us the solutions x = 0 and x = -13.5. Substituting these values back into y = 4.5x gives us the corresponding y-values: y = 0 and y = -60.75. So, the critical points are (0, 0) and (-13.5, -60.75). To determine the nature of these points, we need to find the second partial derivatives of f: f_xx = 6x, f_yy = -2, f_xy = f_yx = 9. The determinant of the Hessian matrix, D = f_xx*f_yy - (f_xy)², will tell us the nature of the points. If D > 0 and f_xx > 0, the point is a local minimum. If D > 0 and f_xx < 0, the point is a local maximum. If D < 0, the point is a saddle point. For the point (0, 0), D = 6*0*(-2) - (9)² = -81 < 0, so this point is a saddle point. For the point (-13.5, -60.75), D = 6*(-13.5)*(-2) - (9)² = 162 - 81 = 81 > 0, and f_xx = 6*(-13.5) < 0, so this point is a local maximum. So, to complete the sentences: • the left-most stationary point (the one with the lower value of x) is located at x = -13.5 and y = -60.75 and is a local maximum point. • the right-most stationary point (the one with the higher value of x) is located at x = 0 and y = 0 and is a saddle point.

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To find the critical points of the function f(x, y) = x³ - y² + 9xy + 5, 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 = 3x² + 9y. The partial derivative of f with respect to y is f_y = -2y + 9x. Setting these equal to zero gives us the system of equations: 3x² + 9y = 0, -2y + 9x = 0. Solving this system of equations gives us the critical points. From the second equation, we can express y in terms of x: y = 4.5x. Substituting this into the first equation gives us 3x² + 9*4.5x = 0, which simplifies to 3x² + 40.5x = 0. This can be factored to 3x(x + 13.5) = 0, giving us the solutions x = 0 and x = -13.5. Substituting these values back into y = 4.5x gives us the corresponding y-values: y = 0 and y = -60.75. So, the critical points are (0, 0) and (-13.5, -60.75). To determine the nature of these points, we need to find the second partial derivatives of f: f_xx = 6x, f_yy = -2, f_xy = f_yx = 9. The determinant of the Hessian matrix, D = f_xx*f_yy - (f_xy)², will tell us the nature of the points. If D > 0 and f_xx > 0, the point is a local minimum. If D > 0 and f_xx < 0, the point is a local maximum. If D < 0, the point is a saddle point. For the point (0, 0), D = 6*0*(-2) - (9)² = -81 < 0, so this point is a saddle point. For the point (-13.5, -60.75), D = 6*(-13.5)*(-2) - (9)² = 162 - 81 = 81 > 0, and f_xx = 6*(-13.5) < 0, so this point is a local maximum. So, to complete the sentences: • the left-most stationary point (the one with the lower value of x) is located at x = -13.5 and y = -60.75 and is a local maximum point. • the right-most stationary point (the one with the higher value of x) is located at x = 0 and y = 0 and is a saddle point.

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