Real Analysis | Where are extreme values?

[@jimallysonnevado3973]

The contrapositive of the theorem states that if f'(c) does not equal to 0 for all c in (a,b) and if f is differentiable on (a,b) then f does not attain maximum or minimum on (a,b) but extreme value theorem (real analysis version) states that there exist a maximum and minimum on [a,b] hence the maximum or minimum is necessarily on the endpoints a or b. We can take also the contrapositive in a different way to arrive at the second conclusion, if f'(c) does not equal to zero for all c in (a,b) but attains maximum or minimum at c in (a,b) then f is not differentiable on c ( f'(c) does not exist ).

[@fenilpatel417]

Please make a course on multi-variable calculus as it is often taught along with real analysis

[@quantabot1165]

I thought you were done with real analysis, I am so excited to learn more from you Thank you

[@goodplacetostop2973]

8:23

[@carstenhaese]

If f(c) is a minimum, set g=-f, so g(c) is a maximum. Apply the proven result to g and you get f'(c) = -g'(c) = -0 = 0.

[@CM63_France]

I agree.

[@CM63_France]

Hi, 2:49 : if f is continous, and it is, because it is diffrentiable, so this maximum (minimum) exists, we don't need to suppose that. 8:04 : to prove the minimum case, just take -f instead of f . For fun: 2 "let's go ahead and", 3 "great", including 1 "ok, great", and the final "and that's a good place to stop" The next time I will count the "the next thing I want to {notice|do}"'s .

[@MichaelPennMath]

We do need to suppose there is a maximum (minimum) as we are working over an open interval in this set-up and not a closed one.

[@tgx3529]

the continous function has always maximum and minimum on closed interval. There is open interval. Open interval is too (-infinity,infinity).Function y=x there has any maximum, this function is definined on R

[@CM63_France]

Ok, thanks.

[@tomatrix7525]

@@CM63_France if you view aroind 5:20 you see that he uses the fact that c is a maximum and thus xn and yn less than it, depsite y greater than c and in a,b. If he hadn’t supposed a maximum existed on a,b then there wouldn’t necessarily be such a c because a greater potential maximum can always be found if f was increasing, since the closer we are to b , f(x) is larger and thus there exists no solid maximum c.

[@joshhickman77]

Wouldn't this be easier just using the standard definition of the derivative, doing a case analysis and using the definition of a limit to say, if the derivative is positive, there exists an epsilon s.t. f(x+epsilon) > f(x), so it's not a maximum, and use the argument on g(x) = f(-x) to get the negative case, and h(x) = -f(x) to get the minimum case?

[@christianaustin782]

At the beginning he mentions that he already proved the extreme value theorem without derivatives. Does anyone know what video that one's in? I can't seem to find it anywhere

[@Benjamingerardbrady]

He proves it towards the end of "Real Analysis | The continuous image of a compact set". kzbin.info/www/bejne/pKK9m2Rom6mZhLc

[@christianaustin782]

@@Benjamingerardbrady Ahh thanks a bunch!

[@roberttelarket4934]

Vote for Michael Penn for President!

[@alexanderskladovski]

ε>

[@ejercicioparatodos1886]

First

[@laurelcreek7]

As an engineer, a user of math, I’m glad somebody is doing rigorous proofs but I have better things to do. A quick sketch would make it obvious that f’(x) has to be zero at a max or min.

[@thecustomer2804]

To be fair, engineers do a lot of Applied Mathematics. Rigorous proofs like the ones found in Real Analysis are oriented more towards people who’re aiming for research into Mathematics.

[@internetbad3575]

Okay, but this is a mathematics channel...

[@CM63_France]

In this case, consider having "better thing to do" than posting in this channel :/

[@axelperezmachado3500]

@ゴゴ Joji Joestar ゴゴ Nice point. Btw, I just wonder. Is that even possible? I mean given the fact that the function is differentiable everywhere then it's derivative function must be continuous everywhere right? So how is it possible for it to be 0 everywhere but positive only at x=0? By the IVT i would expect to be at least a whole open interval in which it is positive

[@marinakalaba2401]

@ゴゴ Joji Joestar ゴゴ would that be a function similar to f(x)=|sgnx|?