Question text Find a critical point [math: [x0,y0]] of the function [math: f(x,y)=e−x+y−2xy]. [math: [x0,y0]=] [input] What is the nature of the critical point? a local minimuma saddle pointinconclusivea local maximum Check Question 1

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Compute partial derivatives: let phi(x,y) = -x + y - 2xy so f = e^{phi}. Then f_x = phi_x e^{phi} = (-1 - 2y)e^{phi}, f_y = phi_y e^{phi} = (1 - 2x)e^{phi}. Set f_x = 0 and f_y = 0. Since e^{phi} > 0 for all x,y, the factors must vanish: -1 - 2y = 0 => y = -1/2, 1 - 2x = 0 => x = 1/2. Thus the critical point is (x0,y0) = (1/2, -1/2). To classify the critical point compute second derivatives. Using phi-derivatives, phi_x = -1 - 2y, phi_y = 1 - 2x, phi_xx = 0, phi_yy = 0, phi_xy = -2. Formulas: f_xx = (phi_xx + phi_x^2) e^{phi}, f_yy = (phi_yy + phi_y^2) e^{phi}, f_xy = (phi_xy + phi_x phi_y) e^{phi}. Evaluate at (1/2,-1/2): phi_x = 0, phi_y = 0, phi_xy = -2, and phi(1/2,-1/2) = -1/2 so f0 = e^{-1/2}. Therefore f_xx(1/2,-1/2) = (0 + 0^2) e^{-1/2} = 0, f_yy(1/2,-1/2) = (0 + 0^2) e^{-1/2} = 0, f_xy(1/2,-1/2) = (-2 + 0·0) e^{-1/2} = -2 e^{-1/2}. Compute the Hessian determinant D = f_xx f_yy - (f_xy)^2 = 0·0 - (-2 e^{-1/2})^2 = -4 e^{-1} < 0. A negative Hessian determinant indicates the quadratic form has mixed signs, so the critical point is a saddle point. Why the other choices are incorrect: local minimum or maximum require D > 0 (and f_xx positive for min or negative for max); here D < 0 so neither applies. "Inconclusive" is wrong because D gives a definitive classification (negative → saddle). Conclusion: the critical point is (1/2, -1/2) and it is a saddle point.

Question text Find a critical point [math: [x0,y0]] of the function [math: f(x,y)=e−x+y−2xy]. [math: [x0,y0]=] [input] What is the nature of the critical point? a local minimuma saddle pointinconclusivea local maximum Check Question 1Multiple fill-in-the-blank

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