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

The most beautiful way is to notice that the function is convex on the positives, which is easily proven. Thus it only has one minimum. Next (now comes the beautiful part) it is invariant under the transformation x->1/x. Thus, also the minimum has to be invariant (since there is only one). This finally leads to x=1.

AM GM is the best until u study calculus

It looks like my hands are tied huh😂

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

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

In Spanish notation, the parentheses are superfluous and usually omitted.

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

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

The frogs in my garden pond solved it like this: Observe that (√x - 1/√x)² = x - 2 + 1/x Hence x + 1/x = (√x - 1/√x)² + 2 This quadratic form has a minimum value of 2 when (√x - 1/√x) = 0 which happens when x = 1

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

First, x + 1/x -> -infinty as x -> 0 from the left, so I'm going to assume we are minimizing over positive x. Notably, when x > 0, y = x + 1/x > 0 as well. Doing some rearranging, we get that x^2 - yx + 1 = 0. We want to minimize y, so we choose the smallest positive y such that there exists a positive x which satisfies the equality. Since it is a quadratic, we know there exists a real x which satisfies it whenever the discriminant is non-negative. Here, the discriminant is y^2 - 4, and the smallest positive y that keeps it non-negative is the one that makes it zero, i.e. y = 2. Plugging this back into the quadratic, we can see that x =1 > 0, so this is a valid choice and thus it is the minimum.

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

I did something somewhat similar to method 3, but not quite as rigorous: Let x + 1/x = b and multiply by x. Then you get a quadratic x^2 - bx + 1 = 0 This has a repeated root at stationary points (graphically, this is obvious), and we get b = 2 as the relevant minimum.

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

method 3 is mvp for me... though I would approach in a bit different way. 1) Suppose we have positive a and x, such that a=1/x+x 2) Multiply it by x and gather all terms on one side: x^2-a*x+1=0 3) we got quadratic equation but we already know the x (by assumption 1), so we know there is at least one root. So the determinant must be nonnegative D = a^2-4*1*1 = a^2-4 >= 0 4) and so a>=2 In the end we got: if a=1/x+x then a>=2. I like this method more, since we find "2" and not consider it from the begining

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!

Let f(x) = x + 1/x | x € R\{0} Notice that f(1/x) = 1/x + x = f(x) From this we get two cnadidates for extremum points, x=+-1 , sinc for example for all x>1, f(x) > f(1), then necessarily for all 0 2 = f(1), so our guess will be that 1 is a minimum point Let x>0 We will try to find a solution to the inequality f(x) > f(1) x + 1/x > 2 x² + 1 > 2x x² - 2x + 1 > 0 (x-1)² > 0 This inequality holds for all x=/1 in the domain ==> x = 1 is a minimum of the function, solved using algebra only

You could also use graph where take f(x) = max(x,1/x) which would give you the minimum point of the graph at (1,2)

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.

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.

MY SOLUTION (SPOILERS!) I have the stinking suspicion that the answer is gonna be f(1) = 2 only considering the positive domain, of course pf: f(1/x) = f(x) which means I can just focus my attention on x >= 1 \begin{SKIPPABLE} split f(x) into x and 1/x. Notice that for all x >= 1, 0 < x = 1, x < f(x) = 2. \end{SKIPPABLE} Show that f(x) is increasing along the interval [1, +inf) Assume we have some x >= 1 and some positive offset s > 0. Then the goal is to show that f(x) < f(x+s). We have the following chain: f(x) = x + 1/x = (x+s)(x + 1/x)/(x+s) = (x^2 + 1 + sx + s/x)/(x+s) = (x^2 + 1 + sx)/(x+s) + (s/(x+s))(1/x)

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

Assuming real numbers, A + B is always greater or equal to 2sqrt(AB) because (sqrt (A) - sqrt (B))^2 must either be positive or zero; in other words: (sqrt (A) - sqrt (B))^2 not less than 0; Multiply out and add 2 sqrt(A)sqrt(B) to both sides We get: A + B not less than 2 sqrt(A)sqrt (B) In this case the minimum is 2sqrt(x* 1/x) = 2 which is our answer.

If x is positive, then noting that f(x) = f(1/x) implies an extrema at x = 1/x --> x =1 --> f(1) = 2. endpoints imply a minimum since f(0+) = f(+infty) = +infty.

Complete the square on sqrt(x)^2 + (1/sqrt(x))^2 = (sqrt(x) + 1/sqrt(x) )^2 -2 = (sqrt(x) - 1/sqrt(x) )^2 + 2

over which interval

I've done a lot of inequality problems for fun with my colleague and I've found that they're always either AMGM, Jensen, or Cauchy-Schwarz

Haven’t watched the video yet but it’s literally: AM-GM: x+1/x >= 2sqrt(x*1/x). So x + 1/x >= 2 meaning the it’s smaller value is 2

My rule of thumb is "the minimum value of x+f(x) is reached when x=f(x)",not sure if this works for all f(x), and may need AM GM for multiple terms

Well, It doesn't prove it's a minimum. You've just found some value, that doesn't exceed the function. You could've argued the same thing for 1, for that matter.

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.

You an simply factor it as (√x-(1/√x))² +2 minimum value of a whole square quantity is 0 so that leaves us with minimum value 2.

negative infinity

f(x) = 2cosh(lnx) for x>0 and cosh has a minimum of 1, so f(x) > = 2 and = 2 when x = 1

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....

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.

Minimum at x approaches -infinity.

When you realize positivity of x^2 in real world

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

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

0:03 Rosana Rodríguez-López approves. Maybe.

Why didnt a get a notification?

So this is the 911th video.

Are you the guy who made that video 54 years ago

Coolio

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...

with calculus you'll also find that the function is minimised at at x=1 and the min is f(1)=2 after taking the second derivative

Maid outfit for 400K?

You should use calculus, that s all