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

$$\mathrm{rM}(0,0,1)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

Campbell University

Oregon State University

Harvey Mudd College

Boston College

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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Find and classify, using t…

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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 E. To the power of negative two, X squared minus three Y squared. To begin, we need to find the critical points. So those will be where the gradient will equal the zero vector. The first element of the gradient is going to be negative four X times E. To the power of negative two, X squared minus three, Y squared. The second element is going to be negative six Y. E. To the power of negative two, X squared minus three Y squared. Clearly the only point which will make that equal to zero is the zero. So now that we have our critical point, we can proceed with applying our testing. So our second partial derivative with respect to X is going to give us four X squared minus one times uh four. Actually, so I should put a four out front there four times four X squared minus one times E. To the power of negative two X squared minus three Y squared the second partial. Why will equal six times six? Y squared minus one times E. To the power of negative two X squared minus three Y squared. And the mixed partial. It's going to equal 24 X Y E. To the power of negative two X squared minus three Y squared. Which then means that our d of x y function is going to equal -24 times six Y squared plus four, X squared minus one. After our uh sorry, times E to the power of negative four, X squared minus six Y squared. Where I've simplified this off screen, Uh then we just need to plug in our values of X&Y equals 00. We get a result of 24 now, one moment here actually. So we can see that that is greater than zero. And our second partial derivative with respect to X evaluated at 00, gives us a value of negative four. So based on our second partial derivative test, that is going to be a relative maximum.

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