Question

Intermediate Value Theorem numero dos

Answer 1

Use the IVT to show that there is a root of the given equation in the specified interval.
\[x^4 + x - 3 = 0, (1, 2)\]

Answer 2

These are easy, Abbles. Plug in x=1, then x=2.

Answer 3

I was thinking about plugging in 1 and 2, but since the interval uses parentheses instead of brackets I wasn't sure...

Answer 4

LOL okay :D

Answer 5

Both do not equal zero, so there is no solution in the interval?

Answer 6

Actually... 1 gives me -1 and 2 gives 15... 0 is between there, so there would be a solution

Answer 7

Can you get from A to B, with one continuous line, without crossing the x-axis?

Answer 8

:P
I see

Answer 9

Calculus complicated simple problems. okay!

Answer 10

"since the interval uses parentheses instead of brackets I wasn't sure..."

I think that's part of the definition of IVT, that it's an open interval.

Answer 11

Oh, I'm not sure... wouldn't it work with a closed interval too though? If a function is continuous between x and y... it will include all the values from x to y...

Answer 12

http://www.sosmath.com/calculus/limcon/limcon06/limcon06.html

Hmm seems like it is a closed interval. Could just be a typo though.

Answer 13

Here is the definition my teacher gave me:
If f(x) is continuous on a closed interval [a, b] and f(a) doesn't equal f(b), then for every value M between f(a) and f(b) there exists at least one value c (a weird E symbol) (a, b) such that f(x) = M

So would this not work for a closed interval? I'm confused

Answer 14

Yes, so that looks the same.. would it not work on an open interval?

Answer 15

(a weird E symbol)

:D

Answer 16

I don't know what its name is but it means "is an element of". I prefer your description.

Answer 17

Lol

Answer 18

I guess it technically wouldn't work on an open interval. Like in this case, if the value at x=1 was on the x-axis, and the value at x>1 was above it, since then it wouldn't actually have a root... idk.

Answer 19

Hmmm... D:

Answer 20

The question says to show that there is a root in the interval, so I'm assuming there IS one? Or do they want me to prove it doesn't exist?

Answer 21

No, you just show that there is a negative y-value, and a positive y-value, and then by the IVT, there must exist a zero y-value in there. Like if you have one slice of bread, and another slice of bread above it, there must be peanut butter between them.

Answer 22

That's known as the PBT.

Answer 23

"I guess it technically wouldn't work on an open interval."
Isn't (1, 2) an open interval?

Answer 24

Hahaha xD
*almond butter theorem

Answer 25

Yeah but i feel like it may have been a typo.

Okay, the ABT.

Answer 26

I looked at almond butter at Trader Joe's yesterday. Didn't buy any. Also didn't buy peanut butter either, i'd rather get the giant costco sized ones.

Answer 27

Google has some interesting things if you google "intermediate value theorem on an open interval"

Answer 28

Here is the problem in the book.. :/

Answer 29

Also, I had to pass up the almond butter at wholefoods a couple days ago. Outrageously priced.

I'm back to making my own almond butter. A little time consuming to soak/dehydrate/roast the almonds, but it tastes better anyway.

Answer 30

Hey what book is that? Looks familiar.

It was only like $2.50 ish at Trader Joe's, about the same as the PB.

That sounds pretty good... Abbles-made almond butter.

Answer 31

Stewart Differential Calculus (AP Edition)\

Almond butter the same price as PB? No way.

Answer 32

You really think it's a typo? perplexing.

Answer 33

Almond butter and abbles go good together, what can I say.

Answer 34

It might have been a little more, I don't really remember.

I don't know, but i don't think it makes much difference.

I bet they do :D

Answer 35

I think I tutored a girl a couple of years ago who used that book. One time she gave me this delicious Cuban food. She was pretty cool.

Answer 36

Hahahaha xD

So should I assume the book is a typo? I'll ask my teacher when I get the chance, but we have a test monday and she'll be absent :O

Answer 37

It's probably fine and doesn't make a difference, since those functions would all be continuous. It'd only be an issue if a function is continuous on (a, b) but not on [a, b].

In short: i wouldn't even worry about it.

Answer 38

Okay, thanks angel :)

Answer 39

Welcome Pebbles :)