# Homework Help: Calc 1 proof IVT Rolles theorem

1. Oct 24, 2014

### 462chevelle

1. The problem statement, all variables and given/known data
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
Then by the IVT there must be at least one zero between a and b.
f'(x)=2+sin(x)
since f'(x) doesnt = 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 im doing a bad job at explaining this, but this is my first proof for class ever, other than geometry in highschool and i was bad at it. Is there anything terribly wrong or that could be improved upon at all?

Last edited by a moderator: Oct 24, 2014
2. Oct 24, 2014

### Staff: Mentor

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.

Last edited: Oct 24, 2014
3. Oct 24, 2014

### 462chevelle

My confusion is that there is no points where f(a)=f(b) so i dont really know what to do with rolles theorem. Ill rewrite that part with some numbers.

4. Oct 24, 2014

### Staff: Mentor

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

5. Oct 24, 2014

### 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.

6. Oct 24, 2014

### Staff: Mentor

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)?

7. Oct 24, 2014