Second derivative test | MIT 18.02SC Multivariable Calculus, Fall 2010

[@sarahgottemoller1322]

This is something I have been struggling with for the last few days, but was beautifully explained in this video and now I really understand it. Thank you so much!

[@Cineburk]

Holy cow! You must be a calculus wizard. No one in my class could understand this after an hour of confusing exposition. But you taught it in less than 9 minutes and it's crystal clear.

[@MirageScience]

If the discriminant is found to be equal to zero at a c.p. then the test is inconclusive. If the discriminant is greater than zero and A negative then the critical point is a maximum. Most people probably got that second part.

[@Palar_Bear]

Thank you for the video! A few questions: why do you check the value of A to determine whether it's a max or min? Is it possible to check the value of C instead, for example? What happens if A = 0 when you check whether it's a max or min (provided you get AC-B^2 > 0)?

[@erikumble]

If A =0, then it is not possible for AC-B^2 to be positive, since the (B^2) term is >= 0 and it gets subtracted, so if A=0, then the formula can at most give 0 which is not > 0. If the formula gives something greater than 0, then AC - B^2 > 0 means AC > B^2. We check A, but we could also check B. This is because both A and B will be positive or both will be negative. If one of them was positive, and the other negative, then their product is negative which is certainly not greater than B^2. So yes, we could just as easily check A or C to determine if it is a local minimum or maximum, because they have the same sign.

[@danielfhb]

if A=0 It means that 0*C -B^2 = -B^2 so that means that AC-B^2 is always less than 0 it can't be grater than 0 because B^2 is always positive

[@adamv.488]

Here the second derivative evaluated at the critical point equals zero, yet it appears in the denominator in the formula that we used to categorize the kinds of critical points...thoughts on this?

[@3rbdh]

thank you Mr. Joel Lewis, that second Derivative test was very helpful, our instructor don't allow us to use calculator though.

[@connyberggren3790]

One issue would be when AC = B^2 then AC-B^2 = 0. how do you determine what kind of point it would be then?

[@explorador26]

gracias.

[@choicefresh]

Why does this test work? It'd be nice to understand how it was derived rather than just memorizing the cases.

[@erikumble]

Here, this might help: www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/reasoning-behind-the-second-partial-derivative-test

[@Palar_Bear]

Never mind my last question, I saw, just as i posted the comment that it's impossible for that to happen.

[@misspeckpeck]

what if AC-B^2 > 0 and A=0?

[@jpnewshazaribagh8130]

Bro then the AC-B^2 thingy won't be more than 0. Unless B is imaginary. 🤔.

[@DanNCy24]

super helpful ty

[@choicefresh]

Thanks for the video though!

[@bearjibster]

love your hair man. love your video

[@RapMastaZ]

helpful, thanks!

[@DANGJOS]

very helpful!!