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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=2 x^{4}+2 y^{2}-64 x^{2}+8 y+3$$

$$\mathrm{rm}(\pm 4,-2,-517) \mathrm{sp}(0,-2,-5)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

Campbell University

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University of Michigan - Ann Arbor

University of Nottingham

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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uh huh. For this problem we are asked to find and classify using the second partial derivative. Test the critical points of the function shown below. So our first step here is to take the gradient of the function and figure out X and Y such that the gradient will equal zero. So the first element by taking the partial derivative with respect to X is going to be eight X cubed minus 128 X. The second element is going to be four, Y plus eight. And we want to solve such that the gradient equals the zero vector. Now we're going to have a few different possible solutions here particularly get clearly that why must equal negative too? And we get a few different values for X. Specifically negative four, zero and positive for So we have our critical points are Uh 0 -2 And Plus or -4 -2. So we next proceed with our test. So first thing that I will do is take the second partial derivatives with respect to X and life. The second partial derivative with respect to X is going to be 24 X squared minus 128. The second partial derivative with respect to why is going to be positive for and the mixed partial is going to be zero. Which means that our d function is going to be our d function evaluated at a generic X. Y. Is going to be four times 24 X squared minus 128. So we can see then that we'll have at D of 0 -2. That will have four times 0 -128. Which we don't really need to worry about calculating that out exactly. We can see that that's great or less than zero, which tells us that will be a saddle point. And then we can actually see that because we are squaring X. We can deal with the plus or minus for at the same time. We'll have four times 24 and 16 -128 and 24 times 16 is going to be significantly greater than 128. So we get d greater than zero at both of those points, combined with clearly the second partial derivative will be greater than zero at both of those points, indicating that we have relative minima at the points Plus or -4 negative two.

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