Proof of existence by I.V.T.

In this video, I showed how to use Intermediate Value Theorem to prove the existence of a number.

[@oraculum_]

You're one of those few people I would actually love to click every video of theirs and immediately give a like without a further thought

[@demongeminix]

Awesome demonstration of the use of the IVT.

[@rceretta]

Beautiful dialect solution!!!! I simply love it!

[@markmajkowski9545]

??? By observation - Every cubic (odd polynomial as well) has at least one real solution. Plus there is a cubic “formula” despite its modern perceived complexity - especially when the x^2 term is 0 which was first solved before the generalized cubic.

[@kragiharp]

Your videos are great, sir. I really appreciate them. Best wishes to you.

[@PrimeNewtons]

Glad you like them!

[@jan-willemreens9010]

... A good day to you Newton despite the bad weather, Didn't I tell you ... BLACKBOARD --> SIR. NEWTON

[@PrimeNewtons]

😊😊😊😊 Thank you!

[@pk2712]

Great demonstration of the use of the intermediate value theorem .

[@Faroshkas]

I love this channel

[@jesusthroughmary]

Every cubic function (in fact every polynomial of any odd degree) must have at least one real root because complex roots come in pairs

[@magefreak9356]

I think you could also say that: f(x) is continuous on all real numbers. As x approaches negative infinity, f(x) approaches negative infinity. As x approaches positive infinity, f(x) approaches positive infinity. I don't know if the intermediate value theorem would apply here as infinity and negative infinity isn't a number. This would also apply to all polynomials which have an odd number for the highest power of x.

[@9adam4]

I don't think you're technically allowed to do that. But what you can do is prove that f(-c) is negative and f(c) is positive, where c is any integer greater than the sum of the absolute values of the coefficients of the polynomial.

[@magefreak9356]

@@9adam4 why do you need the extra restriction of the coefficients?

[@9adam4]

@magefreak9356 The coefficients will determine how far you need to go to assure you have applicable points. For instance, f(x) = x^3 - 1000x^2, you need to get past 1000 before f(-c) is negative.

[@manpreetkhokhar5318]

As per asked question, x^3 + 3 = x; I.e x^3 - x + 3 = 0 Let's define f(x) = x^3 - x + 3 So, question now transforms to checking if f(x) = 0 has any real root. We can clearly examine that, f(-2) = -3 and f(-1) = 3. Since, 0 lies between -3 and 3, so, by IVT, there exists atleast one c between -2 and -1, such that f(c) = 0. This proves, that f has atleast one root.Hence, for some c between -2 and -1, c^3 - c + 3 = 0. Also, by further analyzing the limits at -inf and its increasing,decreasing nature, we can deduce that,it is the only root.

[@abdikadirsalad1572]

Thanks sir . Please make a video on mid term and final exam reviews calculus 1

[@TheTrx3richie]

Good demonstration!

[@tanjilsarker7678]

Thanks for the help!!

[@artandata]

I kept waiting for the numerical real value of the answer!

[@manitubergaming]

U soooo intelligent

[@JayTemple]

I used the Sign Rule (I think that's the name) to show that there is exactly one solution that's negative. As originally stated, the number the problem asks for is a solution of x = x^3 + 3, which works out to x^3 - x + 3 = 0. If x is a solution, let y = -x. Then -y^3 + y + 3 = 0. The polynomial has exactly one sign change, so it has exactly one positive zero, but if y is positive, then x, which is -y, is negative. ETA: Also, I find 10 and -10 to be easier to use than smaller numbers (other than 1). My test here would be (-10)^3 - (-10) + 3 = -1000 + 100 + 3

[@NathanSibali]

Wow

[@sjn7220]

Is there a solution that doesn’t involve guessing?

[@moeberry8226]

Good video brother but you haven’t shown that x^3-x+3 is continuous even though I know it’s a polynomial and it’s continuous on R. Taking the derivative would also be good to show. Because differentiability implies continuity.

[@williamspostoronnim9845]

I like Your English very much!

[@tudorsafir2766]

Isn't that also called Bolzano's Theorem?

[@PrimeNewtons]

Yes, it is!

[@Ahmed-kg2gf]

The question is , "is x real" in order to solve that u need to get the notation for the equation and it is x³+3=x Reenraging the terms X³-x+3=0 Any cubic polynomial with real coefficient must have at least one real solution so the answer is ye , no need to solve

[@BB-sc8ed]

Thank you for saving my ass

[@9adam4]

It's about -1.6717

[@senseof_outrage9390]

Can you post the exact format of the solution. Did you solve the problem yourself... 😅

[@neevhingrajia3822]

Thats a transcendental number right? So how can the a transcendental number be 3 more than its own cube?

[@9adam4]

@neevhingrajia3822 What leads you to believe it's a transcendental number? As you correctly point out, a transcendental number can't be the solution to an equation like this.

[@neevhingrajia3822]

@@9adam4 so are you saying that -1.6717 cubed would be 3 less than -1.6717?

[@9adam4]

@neevhingrajia3822 Yes I am. Put it into the calculator yourself.

[@Noor-kq9ho]

depressed cubics

[@yaronbracha4923]

Who's Dad?

[@holyshit922]

It is not so difficult to calculate x Assume that x is sum of two unknowns,plug in into the equation use binomial expansion , rewrite as system of equations Transform this system of equations to Vieta formulas for quadratic Check if solution of quadratic satisfies system of equations before transformation

[@Harrykesh630]

Using Derivatives would have been a better approach 🤔

[@m.h.6470]

Solution: is there a real solution to x = x³ + 3? x = x³ + 3 |-x³ x - x³ = 3 |therefore x³ < x, which means that, as the result is an integer, x *has* to be negative! (x³ < x would also be possible, if 0 < x < 1, but then the result of x - x³ would not be an integer) Since x³ is growing very fast, x has to be quite small. testing left term assuming x = -1 -1 - (-1)³ = -1 - (-1) = -1 + 1 = 0 testing left term assuming x = -2 -2 - (-2)³ = -2 - (-8) = -2 + 8 = 6 Therefore there is a real solution of x between -1 and -2.

[@AbouTaim-Lille]

X=3x³ , and excluding the trivial solution X=0. We have 3x²= 1 so x=±1/√3

[@colina64]

nice as usual, please try to change your blackboard to a white one or improve the light system for better visualization of your videos, best regards👍

[@sunil.shegaonkar1]

Nearest answer is - 1.672, it is still an approximate. Question remains: is there an exact solution ? Probably not in rational numbers, but may be an irrational one. Oh, That is why they are called irrational number.

[@Taric25]

Why all this proof stuff? Just solve it. x≈-1.6717 or x≈0.83585 ± 1.0469 i. That's it.

[@xyz9250]

Could you post how you solved it ?