In this video, I showed how to use Intermediate Value Theorem to prove the existence of a number.
Жүктеу.....
Пікірлер: 59
@oraculum_6 ай бұрын
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 Жыл бұрын
Awesome demonstration of the use of the IVT.
@markmajkowski95457 ай бұрын
??? 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.
@jesusthroughmaryАй бұрын
Every cubic function (in fact every polynomial of any odd degree) must have at least one real root because complex roots come in pairs
@kragiharp7 ай бұрын
Your videos are great, sir. I really appreciate them. Best wishes to you.
@PrimeNewtons
7 ай бұрын
Glad you like them!
@rceretta4 ай бұрын
Beautiful dialect solution!!!! I simply love it!
@pk27122 ай бұрын
Great demonstration of the use of the intermediate value theorem .
@jan-willemreens9010 Жыл бұрын
... A good day to you Newton despite the bad weather, Didn't I tell you ... BLACKBOARD --> SIR. NEWTON
@PrimeNewtons
Жыл бұрын
😊😊😊😊 Thank you!
@magefreak93565 ай бұрын
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...
@davidcawthorne711516 күн бұрын
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. 😊❤
@Faroshkas7 ай бұрын
I love this channel
@artandata3 ай бұрын
I kept waiting for the numerical real value of the answer!
@TheTrx3richie3 ай бұрын
Good demonstration!
@JayTemple6 ай бұрын
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
@tanjilsarker7678 Жыл бұрын
Thanks for the help!!
@abdikadirsalad1572 Жыл бұрын
Thanks sir . Please make a video on mid term and final exam reviews calculus 1
@Ahmed-kg2gfАй бұрын
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
@davidbrisbane7206Ай бұрын
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.
@manitubergaming9 ай бұрын
U soooo intelligent
@sjn72203 ай бұрын
Is there a solution that doesn’t involve guessing?
@kennethgee200427 күн бұрын
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.
@tudorsafir2766 Жыл бұрын
Isn't that also called Bolzano's Theorem?
@PrimeNewtons
Жыл бұрын
Yes, it is!
@moeberry82264 ай бұрын
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.
@williamspostoronnim98457 ай бұрын
I like Your English very much!
@NathanSibali9 ай бұрын
Wow
@BB-sc8ed11 ай бұрын
Thank you for saving my ass
@holyshit922 Жыл бұрын
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
@Harrykesh6302 ай бұрын
Using Derivatives would have been a better approach 🤔
@m.h.64708 ай бұрын
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.
@9adam45 ай бұрын
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.
@hridikkanjilal46024 күн бұрын
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??
@yaronbracha49232 ай бұрын
Who's Dad?
@AbouTaim-Lille2 ай бұрын
X=3x³ , and excluding the trivial solution X=0. We have 3x²= 1 so x=±1/√3
@Noor-kq9ho7 ай бұрын
depressed cubics
@elreturner12272 күн бұрын
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
@johnconrardy848621 күн бұрын
why don't you why don't you create a tee shirt with your famous saying
@colina648 ай бұрын
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.shegaonkar16 ай бұрын
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.
@Taric255 ай бұрын
Why all this proof stuff? Just solve it. x≈-1.6717 or x≈0.83585 ± 1.0469 i. That's it.
Пікірлер: 59
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 ?