Critical Points Classification B192B2

1. **State the problem:** Find the critical points of the function $$f(x,y) = x^2 + xy^2 - 2x - 1$$ and classify each critical point using the second derivative test. 2. **Find the first partial derivatives:** $$f_x = \frac{\partial}{\partial x}(x^2 + xy^2 - 2x - 1) = 2x + y^2 - 2$$ $$f_y = \frac{\partial}{\partial y}(x^2 + xy^2 - 2x - 1) = 2xy$$ 3. **Set the first derivatives equal to zero to find critical points:** $$2x + y^2 - 2 = 0$$ $$2xy = 0$$ 4. **Solve the system:** From $$2xy = 0$$, either $$x=0$$ or $$y=0$$. - Case 1: $$x=0$$ Substitute into $$2x + y^2 - 2 = 0$$: $$2(0) + y^2 - 2 = 0 \implies y^2 = 2 \implies y = \pm \sqrt{2}$$ Critical points: $$(0, \sqrt{2})$$ and $$(0, -\sqrt{2})$$ - Case 2: $$y=0$$ Substitute into $$2x + y^2 - 2 = 0$$: $$2x + 0 - 2 = 0 \implies 2x = 2 \implies x = 1$$ Critical point: $$(1, 0)$$ 5. **Find the second partial derivatives:** $$f_{xx} = \frac{\partial}{\partial x}(2x + y^2 - 2) = 2$$ $$f_{yy} = \frac{\partial}{\partial y}(2xy) = 2x$$ $$f_{xy} = \frac{\partial}{\partial y}(2x + y^2 - 2) = 2y$$ 6. **Use the second derivative test:** Calculate the discriminant $$D = f_{xx} f_{yy} - (f_{xy})^2$$ at each critical point. - At $$(0, \sqrt{2})$$: $$f_{xx} = 2$$ $$f_{yy} = 2(0) = 0$$ $$f_{xy} = 2(\sqrt{2}) = 2\sqrt{2}$$ $$D = (2)(0) - (2\sqrt{2})^2 = 0 - 8 = -8 < 0$$ Since $$D < 0$$, this point is a **saddle point**. - At $$(0, -\sqrt{2})$$: $$f_{xx} = 2$$ $$f_{yy} = 0$$ $$f_{xy} = 2(-\sqrt{2}) = -2\sqrt{2}$$ $$D = (2)(0) - (-2\sqrt{2})^2 = 0 - 8 = -8 < 0$$ This is also a **saddle point**. - At $$(1, 0)$$: $$f_{xx} = 2$$ $$f_{yy} = 2(1) = 2$$ $$f_{xy} = 2(0) = 0$$ $$D = (2)(2) - 0^2 = 4 > 0$$ Since $$D > 0$$ and $$f_{xx} > 0$$, this point is a **local minimum**. 7. **Final classification:** - $$(0, \sqrt{2})$$ saddle point - $$(0, -\sqrt{2})$$ saddle point - $$(1, 0)$$ local minimum