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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=2 x^{2}-8 x+3 y^{2}+12 y+2$$

$$\mathrm{rm}(2,-2,-18)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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02:03

Find and classify, using t…

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01:04

mhm For this problem we are asked to find and classify using the second partial derivative. Test the critical points of the function F of x Y equals two, X squared minus eight. X plus three, Y squared plus 12 Y plus two. So to begin we want to find where the gradient of the function is going to equal zero. So the first element of the gradient is going to be four x minus eight. That's the partial derivative with respect to X. The second element is going to be 65 plus 12. And we want to solve for X and Y when the gradient will equal the zero vector. Now the solution to that is going to be X equals two. Y equals negative two. So that suggests that we have a critical point to negative two. So now that we have the critical point, we need to proceed to our test which we do by taking the second partial derivative with respect to Y and X. So the second with respect to X is going to be for second partial derivative with respect to why is going to be six. And we can see that the mixed partial derivative is going to be zero. So our d function is going to be four times 6, 24. Which we can clearly see is greater than zero. And we can see that the second partial derivative respect to X is greater than zero. Which means that by the second partial derivative test we know that that excuse me that that means that there is a relative men relative minimum at the point to negative two.

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