Question

Math

Posted 3 months ago

```
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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Answer from Sia

Posted 3 months ago

Solution by Steps

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.

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