The scalar field $f(x, y)=x \ln \left(y^{2}\right)-x$ has a critical point at $(0, \sqrt{e})$.
How does the second partial derivative test classify this critical point?
Choose 1 answer:
(A) Local maximum
(B) Local minimum
(C) Saddle point
(D) The test is inconclusive

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To classify the critical point using the second partial derivative test, we first compute the second partial derivatives of $f(x, y)$. ‖ ‖ step 2 ⋮ The second partial derivatives are $f_{xx}(x, y) = 0$, $f_{yy}(x, y) = \frac{2x}{y^2}$, and $f_{xy}(x, y) = \ln(y^2) - 1$. ‖ ‖ step 3 ⋮ Evaluate the second partial derivatives at the critical point $(0, \sqrt{e})$. We get $f_{xx}(0, \sqrt{e}) = 0$, $f_{yy}(0, \sqrt{e}) = 0$, and $f_{xy}(0, \sqrt{e}) = 1$. ‖ ‖ step 4 ⋮ Calculate the determinant of the Hessian matrix at the critical point, $D = f_{xx}(0, \sqrt{e}) \cdot f_{yy}(0, \sqrt{e}) - [f_{xy}(0, \sqrt{e})]^2$. ‖ ‖ step 5 ⋮ Substituting the values, we find $D = 0 \cdot 0 - (1)^2 = -1$. ‖ ‖ step 6 ⋮ Since $D < 0$, the second partial derivative test indicates that the critical point is a saddle point. ‖ ∻1 Answer∻ ***C*** ∻Key Concept∻ ⚹Second Partial Derivative Test⚹ ∻Explanation∻ ⚹The second partial derivative test uses the determinant of the Hessian matrix to classify critical points. If the determinant is negative, the critical point is a saddle point.⚹◊What is the formula for the second partial derivative test in multivariable calculus to classify critical points?⍭ Generate me a similar question◊

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