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

Awesome demonstration of the use of the IVT.

??? 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.

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

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

@PrimeNewtons

7 ай бұрын

Glad you like them!

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

Great demonstration of the use of the intermediate value theorem .

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

@PrimeNewtons

Жыл бұрын

😊😊😊😊 Thank you!

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

5 ай бұрын

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

5 ай бұрын

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

@9adam4

5 ай бұрын

@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

3 ай бұрын

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.

@user-dc4xy7uk3b

9 сағат бұрын

@@9adam4 yes you can because the graph is continuous and always increasing...

It is real important to remember that the function has to be continuous for IVT to hold or at least continuous over the interval a b. And to always wear a cool hat when doing mathematics. 😊❤

I love this channel

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

Good demonstration!

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

Thanks for the help!!

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

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

Cardano's cubic formula always find a real solution to a cubic equation, so at least one such c exists and we know f(c) = 0.

U soooo intelligent

Is there a solution that doesn’t involve guessing?

well the question changed a little from the thumb nail to the question on the board. This one feels like a no, but it is the proof they want.

Isn't that also called Bolzano's Theorem?

@PrimeNewtons

Жыл бұрын

Yes, it is!

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.

I like Your English very much!

Wow

Thank you for saving my ass

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

Using Derivatives would have been a better approach 🤔

Solution: is there a real solution to x = x³ + 3? x = x³ + 3 |-x³ x - x³ = 3 |therefore x³ (x³ 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.

It's about -1.6717

@senseof_outrage9390

4 ай бұрын

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

@neevhingrajia3822

3 ай бұрын

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

@9adam4

3 ай бұрын

@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

3 ай бұрын

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

@9adam4

3 ай бұрын

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

All negative number is greater than it's cube

@sinexitoalmiedo

Күн бұрын

False because (-(1/2))^3=-1/8 but -1/8 is greater than -1/2

@hridikkanjilal460

Күн бұрын

@@sinexitoalmiedo I forgot to add 'all negative ' non fractions of decimals are greater than their squres

@sinexitoalmiedo

9 сағат бұрын

@@hridikkanjilal460 what??

Who's Dad?

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

depressed cubics

I did x^3 -x =-3 x(x^2 -1)=-3 x((x+1)(x-1))=-3 x=-3, x=-4, and x=-2 and not one is right

why don't you why don't you create a tee shirt with your famous saying

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👍

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.

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

@xyz9250

4 ай бұрын

Could you post how you solved it ?