Intermediate Value Theorum - WTF?

[frozenstellar]

howdy brainiacs and math gods (from my reading in this forum, there are quite a few of you here!). Having my first year adv math lectures fall behind schedule, we've rushed the last part of the final lectures given before easter break, so i have a few questions that i can't get answered by the math faculty over the easter break, so i turn to you guys.

The intermediate value theorum, i dont understand what it is. he's stated that there are essentially three parts to it

1. f is bounded on the interval [a,b]
2. f attains a maximum M and a minimum m in the interval [a,b]
3. f takes every value in [m,M] at least once for the interval x E [a,b]

i understand what each three parts mean - but is there a method of using this theorum? i submitted homework earlier today to a question (homework doesnt get marked, just get feedback on errors)

show that equation x + sin(x) = 1 has atleast one solution in the interval [0, pi/6]

i didnt show any calculations, just statements regarding what that curve does over [0,pi/6]. i know people dont like discussing homework - but i wont get feedback on this until the 21st april, and the midsemester is on the 23rd, so if anyone could clarify what exactly i am supposed to do with this theorum, it'd be absolutely awesome.

Cheers

[KnifeMissile]

No, don't worry about this being homework. You're genuinely confused and I'm certain that all your confusion will be immediately dispelled and replaced by wonderment and understanding if you simply had someone show you an example.

Besides, I'm the only one on this board who dislikes doing other people's homework!

Anyway, your description of the Intermediate Value Theorem, or their properties (you called them "parts"), is kind of vague. It's hard to tell what you know about it so I will just tell you what I know and you can work that into your course (it really should all be the same).

You can find some good definitions of the Intermediate Value Theorem right here. I'll tell you the one I was taught (by Michael Spivak) but you should understand that they're all equivalent (I will even leave it as an exercise for you to prove).

The Intermediate Value Theorem states that if f is continuous on [a,b] and f(a) < 0 < f(b), then there is some x in [a,b] such that f(x) = 0.

So, lets use this theorem to prove that x + sin(x) = 1 has, at least, one solution.

Let f(t) = t + sin(t) - 1 in the interval [0,&pi/6].
f(0) = -1 < 0.
f(&pi/6) = &pi/6 + 0.5 > 0.
By the Intermediate Value Theorem, there is an x in [0,&pi/6] such that f(x) = 0.
But f(x) = x + sin(x) - 1.
So x + sin(x) - 1 = 0.
Equivalently, x + sin(x) = 1.
QED...

[Poloboy]

For a simpler explanation, because it sounds like that's what you want, I'll give you this:

If you have a function on a given interval, no matter how straight or how squiggly, that has some maximum and some minimum, it HAS to go through every single point in between the maximum and the minimum.

That means that If you have a function with a minimum of -2, and a maximum of 3, and the function satisifies the requirements of the theorem (is continuous, bounded on the closed interval), you know for certain that the function passes through every known point between the values -2 and 3.

What does this mean to you? If your teacher asks you on a test to prove that on the interval [-1,1], the function y=x has a value of y=0 at some point, you use the theorem to prove it. Looking at the graph, it's quite clear that it's a straight line, and that it passes through y=0, but in math you have to cite a theorem if you're going to make a claim like that.

Don't try to look too deep into it. It's as simple as it sounds. It's just a really basic theorem that describes common-sense behaviour of functions, with the purpose of introducing students to referring to known theorems to make claims/proofs.

[Yakk]

As a general guild, the proof of a theorem often gives hints asto how it can be used. Try reading through the proof of the IVT in order to find an answer.

A solution using your version of the IVT would be something like:
Let f(x) = x + sin(x)
We are required to prove 1 is an element of f([0, π/6]). (f[0,π/6]) is the image under f of [0, π/6] See Footnote1 for a more fomal treatment of "image under").
Let A := f([0, π/6])

Let m be the minimium value of A. Let M be the maximium value of A. (note A = [m,M], by IVT(2) and IVT(3).)

Now, f(π/6) = 1.5. Thus, M >= 1.5.
f(0) = 0. This, m <= 0.

From this we know [0, 1.5] is a subset of [m, M].

Now, A = [m,M] (by the intermediate value theorem, part 2). 1 is an element of [0,1.5], which is a subset of A, which is f([0,π/6]). By the definition of f([0,π/6]), there exists an x in [0, π/6] such that f(x) = 1.

As an aside, we never discovered what m and M are: we only found an inequality restricting their values. We don't need to know what their value is in order to find them useful.

Footnotes:
1: If g is a function and B is a set, then "the image under g of B" is defined to be all the values {g(x) such that x is an element of B}. This is denoted g(B).

In other words, y is in g(B) if and only if there exists an x in B such that g(x) = y.

A property of images under functions is Monotonicity: if B is a subset of C, then g(B) is a subset of g(C). (not nessicarially a strict subset, but a subset). Ie, if y is an element of g(B), y is an element of g(C).

[frozenstellar]

thanks for the replies lads, this appears to make a lot more sense now (note: just finished 14hr shift, and its 2am), but alas, if it sinks in now, it'll sink in tomorrow.

i'm not going to try and attempt math now, but tomorrow i'm going to sit down having read these posts, and i think there was 2 other IVT proofs in the course manual. is it cool if i post the proofs to these q's tomorrow and have you's look over them?

again, thanks so much

[frozenstellar]

i've attempted two questions i had, and if someone could post any feedback, it'd be great.

1. If f(x) = x^3 - 5x^2 + 7x - 9, prove there is a real number c such that f(c) = 100.

let g(x) = x^3 - 5x^2 + 7x - 109 where g(c) = 0.

when x = 0, g(0) = -109
when x = 10, g(10) = 461

let x E [0,10] (did this as there was no interval restiction stated in the question)

g(0) < g(x) < g(10)
-109 < g(x) < 461

as g(x) is continuous over x E [0,10], and the min and max over this domain are -109 and 461 respectively, by the IVT the function must attain every value in [-109, 461] atleast once.

so g(c) = 0
and therefor f(c) = 100, where c is a real number

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2. x^5 - 3x^4 - 2x^3 - x + 1 = 0 has atleast one solution between 0 and 1.

let f(x) = x^5 - 3x^4 - 2x^3 - x + 1, where x E [0,1]

f(0) = 1
f(10) = -4

so..
f(1) < f(x) < f(0)
-4 < f(x) < 1

as f(x) has a min and max of -4 and 1 respectively for x E [0,1], the function attains every value between [-4,1] atleast once as f(x) is a continuous function over the interval x E [0,1]

therefor f(x) = x^5 - 3x^4 - 2x^3 - x + 1 has atleast one solution over x E [0,1]

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any thoughts on where i'm not clear in the proof? the odd part was in the first one, there was no interval restriction. did i do right in restricting it to a given interval? otherwise i cant see any method of showing it.

[Yakk]

frozenstellar, I don't like the style of your proofs. Maybe your markers will...

Yes, in order to use the IVT, you have to pick an interval for question 1. You caught that. But:

let x E [0,10] (did this as there was no interval restiction stated in the question)

g(0) < g(x) < g(10)
-109 < g(x) < 461

This is either meaningless or wrong. Just because x is in [0,10] doesn't mean g(x) is in [g(0),g(10)]. And that is what you said in the above 3 lines.

A proof in math isn't just about reaching the end: it is about getting to the end properly. Especially when you are told where you are supposed to go. =)

There is the same problem in your second "proof".

[iccky]

Originally posted by frozenstellar
1. If f(x) = x^3 - 5x^2 + 7x - 9, prove there is a real number c such that f(c) = 100.

let g(x) = x^3 - 5x^2 + 7x - 109 where g(c) = 0.

when x = 0, g(0) = -109
when x = 10, g(10) = 461

let x E [0,10] (did this as there was no interval restiction stated in the question)

g(0) < g(x) < g(10)
-109 < g(x) < 461

as g(x) is continuous over x E [0,10], and the min and max over this domain are -109 and 461 respectively, by the IVT the function must attain every value in [-109, 461] atleast once.

so g(c) = 0
and therefor f(c) = 100, where c is a real number

As Yak said, not quite right on number 1. The easiest way to do this, if I remember correctly, is to choose an interval, find teh min and max on that interval and then state that G(x) is between that min and max (this is true, whereas saying g(x) is between 0 and 10 because x is between 0 and 10 is not necessarily true).

Its been awhile though.

[KnifeMissile]

I will reword your "proofs" so that they will be correct. However, I must use my version of the Intermediate Value Theorem because your still makes no sense to me (for instance, 2. appears to be the conclusion of the Extreme Value Theorem...).

Originally posted by frozenstellar
i've attempted two questions i had, and if someone could post any feedback, it'd be great.

1. If f(x) = x^3 - 5x^2 + 7x - 9, prove there is a real number c such that f(c) = 100.

let g(x) = x^3 - 5x^2 + 7x - 109 where g(c) = 0.

when x = 0, g(0) = -109
when x = 10, g(10) = 461

This is the part you got right. You picked a good interval to work with and this was something you needed to do if you wanted to use the Intermediate Value Theorem. However, as Yakk pointed out, you made a false conclusion and then tried to build off of it. The rest of your proof should have looked something like this:

Now, because g(0) < 0, g(10) > 0, and g(x) is continuous, we know, from the Intermediate Value Theroem, that there exists a c in [0,10] such that g(c) = 0.

But g(c) = c^3 - 5c^2 + 7c - 109.
So, c^3 - 5c^2 + 7c - 109 = 0,
or c^3 - 5c^2 + 7c - 9 = 100.
QED.

The key here was to use the Intermediate Value Theorem to prove your conclusion.

-----------------------------------

Your second part was similarly flawed but I'll leave it to you to correct it. I'm sure you know how!