Find the local maximum and minimum values and saddle...

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.
$f(x, y)=x^{3}-3 x+3 x y^{2}$

Key Concepts

Partial Derivatives and Critical Points

In multivariable functions, finding the points where the function could have a local maximum, minimum, or saddle point involves first computing the partial derivatives with respect to each variable and setting them equal to zero. These points, where all first-order partial derivatives vanish, are known as the critical points, and they are the candidates for local extrema or saddle points.

Hessian Matrix and Second Derivative Test

Once critical points are determined, the Hessian matrix, which consists of all second-order partial derivatives, is used to classify them. By evaluating the Hessian at each critical point, the second derivative test helps determine if a point is a local minimum, a local maximum, or a saddle point based on the definiteness of the matrix.

Saddle Points

A saddle point is a type of critical point where the function does not exhibit local extremum behavior but instead has a surface that curves upward in one direction and downward in another. The presence of saddle points is significant in understanding the overall topography of the function's graph.

Graphical Visualization

Graphical visualization aids in comprehending the complex behavior of functions of two variables by depicting features like local extrema and saddle points. Using three-dimensional plotting or contour plots allows one to view and verify the analytical classification of critical points, ensuring that all important aspects of the function’s behavior are captured.

Transcript

00:01 Here we are given a function fxy equals x cubed minus 3x plus 3xy squared and we are asked to find the local maximum and minimum values as well as several points.

00:19 And so first we have to find where the critical points are, which we can do by setting second derivatives equal to zero.

00:27 So first let's take the x partial of this to get 3x squared minus 3 .3.

00:32 Plus 3y squared, and we can also take the y partial to get just 6xy.

00:45 And now we look for points that make these values equal to 0.

00:51 And so if we said 0 equal to 3x squared minus 3 plus 3y squared, that's also 1 equals x squared plus y squared.

01:03 And so this is saying that any points on the circle with radius one do have an x partial of zero, but we also need the y partial to be zero.

01:14 And so we say zero equals six xy.

01:20 This tells us that either x or y must be zero.

01:28 So now we know that all of the critical points are points lying on the circle of radius 1 where at least one of the x or y coordinates are 0.

01:36 And so if you think about it intuitively, that gives us kind of the four, if we have our circle there and our axes there, that gives us these four critical points here.

01:51 But the values plus or minus one and zero and zero plus or minus one.

01:59 And so we have to perform the second derivative test on each of these.

02:03 First, we'll go ahead and find general expressions for the second derivatives.

02:09 So fxxx equals 6x, fxy, taking y derivative of this for instance is just 6y, and then fy y y, y partial of this would be 6x...

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