Calculus 1: How to use the intermediate value theorem

Here's an example of how we can use the intermediate value theorem. The cubic equation x^3-3x-6=0 is quite hard to solve but we can use IVT to determine where the solution has to be. Remember the IVT is for continuous functions only. It is useful when we have to determine an interval that has a root to a continuous function.

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[Invalid]

Some nitpicking:
If within an interval [a, b] we have a continuous function f such that: f(a)f(b) < 0 then IVT guarantees there's AT LEAST one solution, not just a solution.

Edit:
In order to guarantee there's ONLY ONE solution we also have to prove that f is monotonous within the interval [a, b].

[harshithsubudhi]

Now u have to prove that it's continuous

[jorgelenny]

Wait so there's a "mean value" theorem and an "intermediate value" theorem? Yeah I'm sticking to Bolzano's for this one

[thewitchking]

For people saying it’s invalid: it doesn’t ask if there are solutions in more than one, it asks which has a solution. If you find that B has a solution feel free to put that choice, you will find that it doesn’t though if you look. His method here is not to disprove any of the other choices, but to correctly choose one choice, by IVT which happens to be fairly quick here

[vitalsbat]

bprp teaches me more than what my teacher teaches (and i understand so I am happy)

[vonakakkola]

you even can prove this is the only real solution because if you see in which points the derivative is equal to zero (in -1 and 1), you have both negative value of the function (y=-8 and -4 in theese points), so the function is decreasing between theese points but in nowhere of the interval the function goes to zero

sorry for the english

[GodbornNoven]

I got it correct! nice!
I guessed it would be answer d because other answers are unlikely tbh.

[JonathanMandrake]

Question: we no have proven that in the interval (2, 3) there is a solution for f(x)=0. But couldn't possibly even 2 of the other intervals contain a solution of this equation? It's quite similar to x^2- 1/4 being 3/4 at 1 and -1 but being 0 at x=+-1/2, so even though we have a solution, the intermediate value theorem is useless

[Manluigi]

with linear recursion sequences we Dn+2= 3 Dn + 6Dn-1 with the initial conditions D1=2, D2=2, D3 =3 . The positive solution of cubic equation is p= 2.355

[Silvarx]

Since "which" doesn't imply only one solution, shouldn't a correct working out include checking f(-1) and f(0) for the bounds of (A) and (B) too?

[chitlitlah]

So answer D clearly contains a solution, but wouldn't you need to take the derivative and make sure there are no local maxima to verify C can't be a solution as well? Or is the intention of the problem that you only use the IVT without taking a derivative?

[GPLB]

If curated carefully, all of these videos are a curriculum in their own

[davidramos]

Which of the following intervals must contain a solution to…..

[genius]

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