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

[sinanmertulucay]

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.

[demongeminix]

Awesome demonstration of the use of the IVT.

[Necrozene]

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!

[magefreak]

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.

[tcoren]

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
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

[markmajkowski]

??? 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.

[pk]

Great demonstration of the use of the intermediate value theorem .

[rceretta]

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

[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

[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 << 0

[davidcawthorne]

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. 😊❤

[davidbrisbane]

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.

[kianushmaleki]

Love the "natural " joke. 😅🤣

[jan-willemreens]

... A good day to you Newton despite the bad weather, Didn't I tell you ... BLACKBOARD --> SIR. NEWTON <-- PRESENTATIONS ... is your trademark, don't forget this! Even MIT still uses this way to teach, no digital devices needed ... By the way my friend, great clear application example of the I.V.T. Thank you again Newton, Jan-W

[michaelbujaki]

The number turns out to be approximately -1.67169988165716.

[artandata]

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

[abdikadirsalad]

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

[davidbrisbane]

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.