??? 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.
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 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.
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
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
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.
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.
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
@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.
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.
??? 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
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. 😊❤
Your videos are great, sir.
I really appreciate them.
Best wishes to you.
Glad you like them!
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...
... 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!
Great demonstration of the use of the intermediate value theorem .
I love this channel
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
I kept waiting for the numerical real value of the answer!
Good demonstration!
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.
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
Thanks sir . Please make a video on mid term and final exam reviews calculus 1
Thanks for the help!!
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.
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.
U soooo intelligent
Is there a solution that doesn’t involve guessing?
Isn't that also called Bolzano's Theorem?
Yes, it is!
I like Your English very much!
Thank you for saving my ass
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.
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
Wow
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.
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
X=3x³ , and excluding the trivial solution X=0. We have 3x²= 1 so x=±1/√3
Using Derivatives would have been a better approach 🤔
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
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👍
Who's Dad?
why don't you
why don't you create a tee shirt with your famous saying
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.
Could you post how you solved it ?