I immediately answered "yes". In my mind I saw the graphs of y = x, y = x^3, and saw there were differences above and below +3, and I did not know exactly where. So, inituitively, I used the IVT. Cheers!
I think it could also be proved like this: x = x^3 + 3 x^3 - x + 3 = 0 let P(x) = x^3 - x + 3 Inspecting the coefficients: 1 ∈ R, -1 ∈ R and 3 ∈ R. Hence, P(x) is a real cubic polynomial. Therefore it either has: a. 3 real solutions (x - R1)(x - R2)(x - R3) b. 1 real and 2 complex solutions (x - c)(x - a - bi)(x - a + bi), a ∈ R and b ∈ R So, there exists at least one real x such that P(x) = 0.
??? 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.
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.
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 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.
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.
You can also use the fundamental theorem of arithmatic, which is arguably more advanced but is a bit more generic and is more fundamental to the question. The idea is to prove every odd degree polynomial with real coefficients has a real root. Every n degree polynomial has exactly n roots, and can be decomposed as p(x)=a(x-x1)(x-x2)...(x-xn), with x1,x2,...xn the roots. Taking the complex conjugate should result in the same polynomial since it's real, so p(x)=a(x-x1*)(x-x2*)...(x-xn*) Equating the roots of the two versions of the polynomial shows that each coefficient is either real or comes in a pair with another coefficient such that each is the other's coefficient. Since it is impossible to arrange an odd number of numbers into pairs at least one of them must be real
@@boguslawszostak1784I mean it depends on what you are allowed to start with. I do not recall a proof of it over the top of my head. For what it's worth it's usually taught & proven before the intermediate value theorem in university, and is IIRC easier to prove, even if it makes less intuitive sense
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 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
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.
That's circular logic. The proof for the derivative of a polynomial goes through proving it is continuous. It's like using L'hopital's rule to find the limit of sinx/x
The question is “Is x real?”. You have answered the question “Can x be real?”. In order to answer the given question, you must rule out option 2, in which two of the possibilities for x are imaginary. For a cubic equation with real coefficients, there are these possibilities: three real solutions, two real solutions (one with multiplicity two), or one real solution and two imaginary solutions. So the answer to the question could be “no”. More work to do. You did, however, answer the question in the thumbnail for the video.
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
@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.
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
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.
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.
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
I feel you, that’s exactly what I just did!
Awesome demonstration of the use of the IVT.
I immediately answered "yes". In my mind I saw the graphs of y = x, y = x^3, and saw there were differences above and below +3, and I did not know exactly where. So, inituitively, I used the IVT. Cheers!
I think it could also be proved like this:
x = x^3 + 3
x^3 - x + 3 = 0
let P(x) = x^3 - x + 3
Inspecting the coefficients: 1 ∈ R, -1 ∈ R and 3 ∈ R. Hence, P(x) is a real cubic polynomial.
Therefore it either has:
a. 3 real solutions (x - R1)(x - R2)(x - R3)
b. 1 real and 2 complex solutions (x - c)(x - a - bi)(x - a + bi), a ∈ R and b ∈ R
So, there exists at least one real x such that P(x) = 0.
Your videos are great, sir.
I really appreciate them.
Best wishes to you.
Glad you like them!
The number turns out to be approximately -1.67169988165716.
??? 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.
well yeah i think this is for more general proofs of this type
... A good day to you Newton despite the bad weather, Didn't I tell you ... BLACKBOARD --> SIR. NEWTON
😊😊😊😊 Thank you!
Beautiful dialect solution!!!! I simply love it!
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.
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.
@@9adam4 why do you need the extra restriction of the coefficients?
@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.
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.
@@9adam4 yes you can because the graph is continuous and always increasing...
Great demonstration of the use of the intermediate value theorem .
I love this channel
Every cubic function (in fact every polynomial of any odd degree) must have at least one real root because complex roots come in pairs
Or if one goes towards plus or minus infinity he x³ term dominates and the value also goes towards the same infinity.
@okaro6595 true, and therefore since it tends to both infinities it must pass through 0 on the way
If the coefficients of a polynomial are complex numbers, it doesn't need to have a real number root
@@김민국-c9f but they are not here..
Excellent!
You can also use the fundamental theorem of arithmatic, which is arguably more advanced but is a bit more generic and is more fundamental to the question.
The idea is to prove every odd degree polynomial with real coefficients has a real root.
Every n degree polynomial has exactly n roots, and can be decomposed as p(x)=a(x-x1)(x-x2)...(x-xn), with x1,x2,...xn the roots.
Taking the complex conjugate should result in the same polynomial since it's real, so p(x)=a(x-x1*)(x-x2*)...(x-xn*)
Equating the roots of the two versions of the polynomial shows that each coefficient is either real or comes in a pair with another coefficient such that each is the other's coefficient. Since it is impossible to arrange an odd number of numbers into pairs at least one of them must be real
And how will you prove this property
@@boguslawszostak1784which property?
@@tcoren1 fundamental theorem of arithmatic
@@boguslawszostak1784I mean it depends on what you are allowed to start with. I do not recall a proof of it over the top of my head.
For what it's worth it's usually taught & proven before the intermediate value theorem in university, and is IIRC easier to prove, even if it makes less intuitive sense
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. 😊❤
Love the "natural " joke. 😅🤣
Thanks sir . Please make a video on mid term and final exam reviews calculus 1
Please can you use the cubic formula to solve this question?
I kept waiting for the numerical real value of the answer!
Good demonstration!
x³ - x + 3 = 0 has one real root and two complex conjugate roots.
In fact, Descartes' Rule of Signs tells us that one negative real root.
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.
Thanks for the help!!
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
what about iteration?
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.
That's circular logic.
The proof for the derivative of a polynomial goes through proving it is continuous.
It's like using L'hopital's rule to find the limit of sinx/x
U soooo intelligent
Is there a solution that doesn’t involve guessing?
Yep
Isn't that also called Bolzano's Theorem?
Yes, it is!
You are Soo good but you like to talk more please make short
The question is “Is x real?”. You have answered the question “Can x be real?”. In order to answer the given question, you must rule out option 2, in which two of the possibilities for x are imaginary. For a cubic equation with real coefficients, there are these possibilities: three real solutions, two real solutions (one with multiplicity two), or one real solution and two imaginary solutions. So the answer to the question could be “no”. More work to do. You did, however, answer the question in the thumbnail for the video.
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
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.
It's about -1.6717
Can you post the exact format of the solution.
Did you solve the problem yourself... 😅
Thats a transcendental number right? So how can the a transcendental number be 3 more than its own cube?
@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.
@@9adam4 so are you saying that -1.6717 cubed would be 3 less than -1.6717?
@neevhingrajia3822
Yes I am. Put it into the calculator yourself.
Who's Dad?
I like Your English very much!
Wow
I just used -10 and 0, these are easy enough
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
x = 1/3 (81/2 - (3 sqrt(717))/2)^(1/3) + (1/2 (27 + sqrt(717)))^(1/3)/3^(2/3)
so exists :p
Using Derivatives would have been a better approach 🤔
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
Thank you for saving my ass
Approximately: -1.67...
depressed cubics
why don't you
why don't you create a tee shirt with your famous saying
Why all this proof stuff? Just solve it. x≈-1.6717 or x≈0.83585 ± 1.0469 i. That's it.
Could you post how you solved it ?
Lol. Do it yourself, this isnt a classroom.
😂
@@xyz9250Sure, plug the coefficients of x³-x+3=0 into the cubic formula. That's it.
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👍
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.
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.
X=3x³ , and excluding the trivial solution X=0. We have 3x²= 1 so x=±1/√3
All negative number is greater than it's cube
False because (-(1/2))^3=-1/8 but -1/8 is greater than -1/2
@@sinexitoalmiedo I forgot to add 'all negative ' non fractions of decimals are greater than their squres
@@hridikkanjilal460 what??
@@sinexitoalmiedo all negative number which are not fractions or decimals are greater than their cubes