Use IVT to show that f(x) exists on interval (0,1)

[macbowes]

Homework Statement

Using IVT, show that 3(\cos(\pi x)) - 3x + 1 exists on interval (0,1)

Homework Equations

None

The Attempt at a Solution

I think I got the right idea, but on my quiz (this was a question I just had on a quiz) I didn't know if I was finished or not.

So when using 0 on the interval (0,1), this is the equation:

3(\cos(\pi 0)) - 3(0) + 1 =

3(\cos(0)) - 3(0) + 1 =

3(1) + 1 =

3 + 1 = 4

Then I used 1 on the interval, which yielded this:

3(\cos(\pi 1)) - 3(1) + 1 =

3(\cos(\pi)) - 3 + 1 =

3(-1) - 3 + 1 =

-3 - 3 + 1 = -5

So I've resulted in -5 and 4, which is large than (0,1), so does that means it exists on that interval? I've got these answers, but I'm not really sure what I do with them. I ended up just writing "squeeze", but when I handed it in he tsk'd at it, so I don't think that's right, lol.

EDIT: I'm trying to fix my post as I made a blatant error that LC pointed out, but what I'm trying to delete (3(1) = 4, lol) keeps appearing in latex in my post even though it's not there when I click the edit button. Anyone know what's up?

 

[LCKurtz]

Homework Statement

Using IVT, show that 3(\cos(\pi x)) - 3x + 1 exists on interval (0,1)

Homework Equations

None

The Attempt at a Solution

I think I got the right idea, but on my quiz (this was a question I just had on a quiz) I didn't know if I was finished or not.

So when using 0 on the interval (0,1), this is the equation:

3(\cos(\pi (0)) - 3(0) + 1 =

3(\cos(0)) - 3(0) + 1 =

3(1) + 1 =

4 + 1 = 5

Does 3*1 really equal 4?

Then I used 1 on the interval, which yielded this:

3(\cos(\pi (1)) - 3(1) + 1 =

3(\cos(\pi)) - 3 + 1 =

3(-1) - 3 + 1 =

-3 - 3 + 1 = -5

So I've resulted in -5 and 5, which is large than (0,1), ...

Huh?? You have shown f(0) = 4 and f(1) = -5.

Try to draw a continuous function on [0,1] that does that and doesn't have any x where f(x) = 0.

 

[macbowes]

Does 3*1 really equal 4?

Well I'm an idiot, LOL! Fixing it now :D.

Huh?? You have shown f(0) = 4 and f(1) = -5.

Try to draw a continuous function on [0,1] that does that and doesn't have any x where f(x) = 0.

Is that what I'm supposed to get? I suppose it's my lack of understand of what IVT really is. I wasn't really sure what an actually answer to this type of question would look like, so I wasn't sure where to go from there.

 

[LCKurtz]

Does 3*1 really equal 4?



Huh?? You have shown f(0) = 4 and f(1) = -5.

Try to draw a continuous function on [0,1] that does that and doesn't have any x where f(x) = 0.

Is that what I'm supposed to get? I suppose it's my lack of understand of what IVT really is. I wasn't really sure what an actually answer to this type of question would look like, so I wasn't sure where to go from there.

The point of the intermediate value theorem is that a continuous function can't jump over the intermediate values. So if it is positive at one point and negative at another, the only way its graph can get from one to the other is to go through 0. So you know there must be a root between them even if you don't know exactly where it is.

 

[macbowes]

The point of the intermediate value theorem is that a continuous function can't jump over the intermediate values. So if it is positive at one point and negative at another, the only way its graph can get from one to the other is to go through 0. So you know there must be a root between them even if you don't know exactly where it is.

Ooooooooooohhhhhh. That clears up a lot! I could do the math, I just didn't know what I was doing. Aside for 3*1 of course, that's devilishly difficult.