Відео

You can imagine an horizontal line with value c that intersects with the function such that you can use the quadratic equation to find all possible c which gives you at the end c²-4>=0 Because we want only real solution. and because we can have two roots for x that means that the horizontal line intersects twice with the function so we choose |c|=2 to get one root for x and knowing that |c| can't be less than 2 we know for sure that the minimum is at x = 1 when c = 2

what the fuck? 1970? 💀💀💀

I assumed that you mean for +x, so I just figured, without any math, any minimum has to have a single x value associated with it, and only x=1 gives that. Every other point with a given y value has x and it's reciprocal as the answer, and so those x can't be a minimum.

Aaaaah its futere

Him using the greek capital chi instead of x is wonderful and terrible. Thanks, I hate it

Never seen this integral. Thanks flammy. Das ist genug fur heite.

kono dia dx

Just plot the function. It will be obvious as x approaches +-1 the function changes directions. 1 is a local minimum. -1 is a local maximum.

Correct me if I’m wrong, but Method 2 doesn’t proof 2 is the minimum, just that it’s a lower bound.

So, lets just assume that minimum exist and it equals m. So there exist x =/= 0 such as: x + 1/x = m we can multiply it by x and just solve for x. We get rather nasty x1 and x2, but we know that extremum of that will be at arithmetic mean of x1 and x2. Luckily the mean is rather nice: mean(x1, x2) = m/2 So we know that the point we are looking for has following form: (m/2; m). So it must be true that: f(m/2) = m m/2 + 1/(m/2) = m which has two nice solution: m = -2 and m = 2 But we know we are looking for a point that has form (m/2, m), so we get two points: (-1, -2) and (1, 2) Problem is I am not really sure how to proof which point is minimum and which point is maximum without calculus.

negative infinity

Utrudniłeś sytuacja jest łatwiejszy sposób do 2n-4=0/+4 2n=4/:2 n=2

*This video's still processing after 54 years later 💀 🗿*

It's easier. The X part gets further from 0 as the absolute value of x increases. The 1/X part gets closer to zero as the absolute value of X increases. They are pulling in separate directions. So, to find the extrema, just set X = 1/X and solve. The answers are 1 and -1. To find which is the max and which is the min, just plug 1 and -1 into the original equation and solve.

Was it about Gauss?

Muirhead: (1, -1) majorizes (0,0) -> sum_sym xy^-1 >= sum_sym 1 = 2 Let y = 1, then sum_sym xy^-1 = x + x^-1 -> x+x^-1>=2 and equality when x = 1. This proof is only valid when x is a nonnegative teal number, but if x could be a negative real number, then the minium value would approach -inf.

Fakeeeee

Hallo Jens, schönes Video. Du bist damit aber ehrlich gesagt unterfordert. Wie wäre es mit einer richtig harten nuss für dich zu knacken? Voila: int_0^infty (x^3)/(exp(x)-1) dx . Das Integral tritt auf, wenn man aus dem Planckschen Strahlungsgesetz das Stefan-Boltzmann-gesetz herleiten möchte. Ich selbst habe es leider nicht lösen können. Abrr traue ich es zu. Grüsse nach Bitterfeld, Johann

Spanish notation would make parentheses redundant when factorials are applied to large expressions. ¡Clearly another W from the superior orthography!

We have to find the minimum of that function without using calculus. In the first step we use calculus to show that the function is always greater or equal to two...

What a coincidence!! Today only, I watched a visual proof of this.. They took a right angle triangle with two legs as x-1/x and 2 respectively. So the third side must be - x+1/x (By Pythagorean theorem). Now we know that hypotenuse must be greater than legs - so x+1/x ≥ 2

Hi Papa Flammy, I've been wondering for quite some time about how I can find a global maximum and minimum of a function? I realize that taking the derivative test only guarentees local extrema points and not global ones. Is it sufficient to say that, if we have f: I-> R, if I is closed then it must have a minimum and maximum by the EVT, therefore if we have an increasing function on some part of the domain, or a decreasing function on some part of the domain, if we then substitute the outermost part of the domain where f is decreasing/increasing, and then compare it to the local extreme points which we get by differentiating our f, see which is bigger and conclude on that? Is this enough? Also, say f: I-> R, I is an open interval, then we do the same approach however we take the limit and now say they would be our supremum and infimum? I have searched wide and far through the internet but I cannot quite find a good solution about finding global min and maxes of functions. Also, have you considered doing videos on complex integration? Things like contour integration I believe need a lot more "Bigger picture videos" then there are on the internet at the moment, I can't quite find a good intuition about understanding how they work. Thank you for your time and hope you have a great day!

You should use calculus, that s all

Easy-peasy, plot the graph and prove that there is no global minimum and one local using the definition. No calculus needed

last time i was this early to a papa flammy video, andrew was being fed in the basement

By using Calculus, it won't come. It doesn't make sense... If you consider negative numbers also (except zero) To understand what I am saying, Take f(x) = x + 1/x So, f'(x) = 1 - 1/x^2 Thereby, the critical points are x = 1,-1 f''(x) = 2/x^3 This gives us, local maxima at x = -1 and local minima at x = 1 with the maximum value being -1 and minimum value being 1....

When you realize positivity of x^2 in real world

Tan u =x F(x)=tan u + cot u tan u = cosec 2u - cot 2u cot u= cosec 2u +cot 2u F(x)=2cosec 2u =>2

No possible answer as the limit in 0- is -∞.

Coolio

idk about minimum but choose negative delta when delta is smol, or very big, either is fine

You forgot the most important way: Start with axiomatic set theory and derive all relevant theorem from scratch

0:03 Rosana Rodríguez-López approves. Maybe.

My first question is, over what domain? It's clear we can make the function arbitrarily small by approaching x=0 from the negative

AM GM is the best until u study calculus

Ich bin immer wieder erstaunt über die Großzügigkeit, die in dieser Gemeinschaft zum Ausdruck kommt. Es ist, als wäre man in eine warme, unterstützende Decke gehüllt.💞

Clear example of work harder not smarter

He must got his shirt from one of the flat earth conferences

WTF tak percaya sejauh itu tahun dahulu

What! Why integration?

1970🔥

I was thinking the same thing 😅…

1970!?

Someone add the formula for binomial expansion 💀

As an A-level Maths student, this was unnecessarily satisfying. 😭

Why is it saying 1969 to me but to everyone else it's 1970

Is that Geoffrey Verity Schofield???

1970 HOW!!!!

This video has in 1970 before youtube released

THIS POSTED SINCE 1970?????? I’M CONFUSED