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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=x y-2 x-3 y+6$$

$$\operatorname{sp}(3,2,0)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

Johns Hopkins University

Harvey Mudd College

University of Nottingham

Boston College

Lectures

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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 shown below. Step one is to find where the gradient of our function will equal zero. First element of our gradient is going to be y minus two. That's the partial derivative with respect to X. Second element is going to be x minus three. Now we can see that for this to equal the zero vector. Clearly we must have y equal to X equal three, having that. Then we start our actual second derivative test to determine the nature of those critical points. You can see that the second partial derivative with respect to X is going to be zero. The second partial derivative with respect to why is going to be zero and are mixed partial derivative is going to equal one. So our d function is going to be just zero minus one squared. So we get D equals negative one, which is less than zero, which tells us that the point 32 is a saddle point.

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