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

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. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x, y) = 2x^3 − 6x + 6xy^2 a.) local maximum value(s): b.) local minimum value(s) c.)saddle point(s) (x, y, f) =

Transcript

00:01 Okay, we want to find a max and min and saddle points of this function.

00:05 Over here, i have the 3d max min test, so we'll know which ones are which.

00:12 So first we're going to find the derivative with respect to x.

00:15 I hope this is right.

00:16 They don't look exactly right on our end, so i'm hoping that that was a minus 6x there.

00:23 Okay, 6x squared minus 6 plus 6 y squared.

00:27 And then f of y is 012 x y oh okay so both of these have to equal 0 for there to be a critical point so either x is 0 or y is 0 and then if x is 0 we get minus 6 plus 6 y squared equals 0 so y squared equals 1 so y equals plus or minus so we got two critical points to check there, 0, 1, and 0 minus 1.

01:09 And if y is 0, then we get 6x squared minus 6 equals 0.

01:16 So x squared is plus and minus 1.

01:18 Oh, i'm sorry, x is plus and minus 1.

01:20 So we got two more points, 1 ,0 and minus 1 0.

01:26 All right, now we need the second partials.

01:28 So f xx is 12x, 12x, fx y is 12x, f x y is 12, take the derivative of f of x with respect to y, 12y, and then f yy -y is 12x.

01:52 So big d is 12x times 12x minus 12y squared.

02:00 Or 144 x squared minus 144 y squared or 144 x squared minus y squared.

02:15 All right, so let's just make us a table here.

02:19 We have our point ab.

02:22 We have the value of big d.

02:29 Then we need the value of fxx, and then we need the type, and then we need f at that point...