How to prove a cubic equation has exactly one root on an interval (IVT, Contradiction, Rolle's Thm)

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@Ninja20704 2 ай бұрын

It would have been helpful to at least state the Rolle’s theorem. For those who don’t know, the theorem states that if a function f is continuous on [p,q], differentiable on (p,q) and f(p)=f(q) (need not be 0), then there exists r in (p,q) such that f’(r)=0 Here we are using [a,b] as our interval, which is a sub-interval of [-2,2] so the continuity and differentiability requirements still hold for [a,b]. Good video nontheless

@LuisFernando-wm5rg 2 ай бұрын

Thx!

@dan-florinchereches4892 2 ай бұрын

I don't think Rollers theorem itself will be very useful but the consequences: A) between 2 consecutive roots of f(X)=0 there will be at least one c such that f'(c)=0 between them B) between two consecutive roots of f'(X)=0 there will be at most 1 root of f(X)=0 Or we could use the consequences of theorem of Laplace stating that if f'(X)

@ghstmn7320 2 ай бұрын

you can also do it using monotony after IVT. f is differentiable as a polynomial function in the interval [-2,2] with a derivaive f'(x)=3x^2 -15. After simple computation its proven that f'(x)

@dlevi67 2 ай бұрын

That (your first reasoning line - it's continuous and strictly decreasing, thus it has no other zero by IVT) is how I approached it too.

@ianfowler9340 2 ай бұрын

After showing there is at least one root in the interval [2,2], I think it is sufficient to show that there are no local max/min in [2,2] as that prevents any more roots. It's a similar argument but easier to present. Am I missing something?

@dlevi67 2 ай бұрын

Don't know - possibly Steve wanted to use Rolle's theorem to show a use case.

@bprpcalculusbasics 2 ай бұрын

I think showing there are no local min/max on [-2, 2] is equivalent to what I did. The reason that I used the Rolle's theorem is because that's just the book's way.... (Single Variable Calculus, by Stewart, sect 4.2) I know it's not the best reason, haha.

@chaosredefined3834 2 ай бұрын

By IVT, we can also show that there is at least one root between -15 and -2. And we can also show that there is at least one root between 2 and 15. Since it's a cubic, by the fundamental theorem of algebra, it has at most 3 real roots. Well, we have 3 real roots. One between -15 and -2, one between -2 and 2, and one between 2 and 15. Thus, there can be no more roots, including in those ranges. So, there is exactly one root in each of those ranges, and thus exactly one root between -2 and 2.

@henrymarkson3758 2 ай бұрын

Isn't the IVT and f'(x)

@ianfowler9340 2 ай бұрын

I think so.

@Grecks75 2 ай бұрын

Yes, it is sufficient. And bprp effectively used that in his proof by contradiction by showing that f'(x) < 0 on the interval (-2, 2). Since the function f is strictly decreasing on this interval, it cannot have more than one zero in it.

@reakreaksf2 2 ай бұрын

Excuse me teacher! How many Chanels do you have I would like to subscribe to them all.😁

@Albaraa 2 ай бұрын

Hey what’s square root of y(x) = x^2 Calculate that then use x=3

@chinnadurai7485 Ай бұрын

Sir why 1+2+3+4+...... has a value of -1

@christoskettenis880 2 ай бұрын

This method is widely used in the Year 12 elective mathematics course in Cyprus. Here's a link to a video of mine using the exact same technique: kzbin.info/www/bejne/m3fXfmuVrLGjrcU Use generated-subs as the video is in Greek

@dan-florinchereches4892 2 ай бұрын

Hi there. Kind of unrelated to the post but I am visiting Paphos is there something nice I shouldn't miss while I am here. Kind regards

@amoghopprasad8286 2 ай бұрын

what if lets say the intervals are different numbers and not have the same absoulte value

@bjornfeuerbacher5514 2 ай бұрын

Doesn't change the way one solves this.

@mr_angry_kiddo2560 2 ай бұрын

What is limit of ((x-2)(x-3)/(x-4)(x-5)) as x tends to 5. Answer:zero(0) . I need mathematical proof 😢

@Neun_owo 2 ай бұрын

The answer shouldn't be 0, as the limit doesn't exist! For a formal proof, we'll have to use Epsilon-Delta proof. but for the sake of simplicity, lets just say that the limit doesn't exist because when x tends to 5 either from left or right side. (x-5) in the denominator becomes incredibly small, so the function shoots up to infinity as it is trying to divide by a very small number. Thus the limit doesn't exist as x tends to 5.

@michaelfaccone5811 2 ай бұрын

@@Neun_owoone small correction, though I agree with your result. The limit as x approaches 5 from the left is actually negative infinity. The limit approaching from the right, as you said, is positive infinity. Since the two limits are different, then as you said the overall limit does not exist.

@Neun_owo 2 ай бұрын

@@michaelfaccone5811 Ty for the correction!

@dlevi67 2 ай бұрын

Is the denominator just (x-4), i.e. the fraction is (x-3)/(x-4), or is the denominator (x-4)(x-5)? If the latter, then the two responses above are correct; the denominator will go to zero as x -> 5. However, I think your exercise has only (x-4) in the denominator, which means you have three factors in the expression: (x-2), (x-3)/(x-4) (x-5) for the limit to be zero, one of those three has to -> 0 for x -> 5 and the other factors need to exist in the same neighbourhood. You can easily demonstrate (with ε-Δ if rigorous proof is required; by inspection and noting that all monomials are continuous and differentiable in R if a less formal proof is adequate) that (x-5) -> 0 if x -> 5, and all the other factors tend to a finite value, thus their product (i.e. the whole expression) tends to 0.