Second partial derivative test example, part 2

[@tejindersingh3602]

at 5:30 i think there is a mistake because for the (sqrt3,1) you put only sqrt3 for fxy which is 6x but it should be (6(sqrt3))^2 so (sqrt3,1) : (0)(-12)- (6(sqrt3))^2 instead of (0)(-12)-(sqrt3)

[@sstream17]

Yes he did forget to multiply x by 6, so both answers for the ±sqrt(3) should be -108. However, this does not change the fact they are saddle points.

[@yonghuiliew8066]

@@sstream17 Thank you for concluding

[@alejrandom6592]

4:34 Grant: "what is 18 times 6?" Me: 6*10+6*8... Grant: "36 times 3"

[@wassimzaki4583]

How about the absolute max and min? How can we know if the local max in this case (at x=0 and y=0) is an absolute max or not?

[@ryanmike9833]

H = det(Hessian_f). Any relevance? We're dealing with functions of two variables, does this second partial derivative test generalize for f(x_1,x_2,...)?

[@zairaner1489]

Yep. In general, you have too look at all the submatrices of the Hessian nxn matrix (meaning the matrices consisting of the first m rows and the first m colummns for a m

[@ryanmike9833]

Ah, very nice. Thanks.

[@sercantor1]

thank you 3blue1brown

[@ryanmike9833]

Also, what about stationary points where chopping with constant x or y yields not a local max/min, but an inflection point. Ex: (0,0) on f(x,y)=y^3-x^2 or f(x,y)=y^3-x^3. Still saddle points?

[@jursamaj]

Yes. See wiki page on saddle point.

[@pratik6447]

How can we make sure that its global maxima/minima not the local maxima/minima??

[@polharbor5664]

at 6:18 Shouldn't - ( - sqrt3)^2 equal positive 3 ??

[@Dezomm]

No, the term ( - sqrt3)^2 becomes positive because you're multiplying two negatives with each other -> (-sqrt(3))(-sqrt(3)) which becomes a positive, and then multiplying with a negative one (-1) in front which makes it negative.

[@alexandrugabrielburea6632]

Thank you bro

[@Thadnill]

what if H turns out positive, but fxx = positive and fyy = negative (or vice versa), is it a maximum or a minimum then?

[@leonardeuler6170]

It can't be If fxx and fyy are of opp. Sign then fxxfyy - fxy^2 has to be negative .coze fxxfyy will be negative and fxy^2 can never be negative

[@alvin3502]

can anyone help me find the critical point of f(x,y) = e^x cosy

[@angelorumlapsus1]

Fχχ*Fyy-Fχy^2= [(e^x)*cosy]*[-(e^x)*cosy] - {[(e^x)*(-siny)]^2}= - [e^(2x)]*[(cosy)^2] - [e^(2x)]*[(siny)^2]= - [e^(2x)]*{[(cosy)^2]+[siny)^2]}= - [e^(2x)]*1 ==> negative for every x (since [e^(2x)] is always possitive) ...soo it's only saddle points.

[@BenlshTracker]

the way this guys speaks is so similar to sal that it is a lil scary :O

[@ConceptualCalculus]

Better than Sal, IMHO.

[@twistedlot]

cool! just a tiny hair away from mentioning the Hessian

[@terryphi]

done!

[@rikthecuber]

Am I the only one being so late?

[@ayberkgormus2959]

nope I am here

[@aminamoudjar4561]

no :p

[@허성진-w5m]

of course not !!!😀