Calc 1 proof IVT Rolles theorem

[462chevelle]

Homework Statement

Use the intermediate value theorem and rolles theorem to prove that the equation has exactly one real solution
2x-2-cos(x)=0

The Attempt at a Solution

Let the interval be [a,b] and let f(a)<0 and f(b)>0
Then by the IVT there must be at least one zero between a and b.
f'(x)=2+sin(x)
since f'(x) doesn't = 0 anywhere and its always >0, therefore f(x) is increasing throughout its entire domain. Therefore f(a) cannot = f(b) anywhere.

I feel like I am doing a bad job at explaining this, but this is my first proof for class ever, other than geometry in high school and i was bad at it. Is there anything terribly wrong or that could be improved upon at all?

 

[Mark44]

Homework Statement

Use the intermediate value theorem and rolles theorem to prove that the equation has exactly one real solution
2x-2-cos(x)=0
3. The Attempt at a Solution
Let the interval be [a,b] and let f(a)<0 and f(b)>0

No, you can't do this. You need to find numbers a and b for which the two inequalities are true. Just try a few values to see if you can get the function to change sign going from one number to the other.

Then by the IVT there must be at least one zero between a and b.
f'(x)=2+sin(x)
since f'(x) doesn't = 0 anywhere and its always >0, therefore f(x) is increasing throughout its entire domain. Therefore f(a) cannot = f(b) anywhere.

I feel like I am doing a bad job at explaining this, but this is my first proof for class ever, other than geometry in high school and i was bad at it. Is there anything terribly wrong or that could be improved upon at all?

 

[462chevelle]

My confusion is that there is no points where f(a)=f(b) so i don't really know what to do with rolles theorem. Ill rewrite that part with some numbers.

 

[Mark44]

Let me change what I said. Use the IVT first. That will show that there is at least one root.

 

[462chevelle]

alright, so if i make the interval [0,2] i get a - then a +. so that would satisfy the IVT. then i use my info that the derivative is always positive so the function is always increasing, therefore only crosses the x-axis once. How do i factor in rolles theorem? or would i have to use contradiction or something? Or could i say, that since there is no interval where f(a)=f(b) then if the function is increasing in one point then it must be increasing throughout the entire domain.

 

[Mark44]

I think you have to use contradiction. From the IVT, you know that there is a zero (around 1.2, BTW). Call this a.

Now suppose that there is another solution b ≠ a, such that f(b) = 0.
What does Rolle's say?
What do you know about f'(x)?

 

[HallsofIvy]

You have already said

since f'(x) doesn't = 0 anywhere and its always >0, therefore f(x) is increasing throughout its entire domain. Therefore f(a) cannot = f(b) anywhere.

So if f(a)= 0 then we cannot have f(b)= 0 for any other b. that is, f(x)= 0 has at most one solution.

As others have said you cannot simply assert

Let the interval be [a,b] and let f(a)<0 and f(b)>0

until you know that there exist values of x where f(x)< 0 and where f(x)> 0.

You are given that f(x)= 2x- 2- cos(x). Further, you know that cos(x) has value only between -1 and 1. Suppose x is some very large, negative, number, say x= -1000000. What can you say about f(x)? Suppose x is some very large, positive, number, say x= 1000000. What can you say about f(x)?