# Mean Value Theorem?

1. Nov 10, 2007

### 2RIP

The problem statement, all variables and given/known data
a. If f is defined on an interval [x,y], its differentiable on open interval (x,y), and f(x)=f(y) then there is a number c in (x,y) where f'(c)=0

b. Does the absolute value of x, |x|, satisfy Rolle's Theorem on [-1, 1]?

The attempt at a solution

For the first one, I'm not really sure.

But for the second, I believe it does not satisfy. Because when x =0 on the interval, it is not differentiable.

2. Nov 10, 2007

### d_leet

Your answer for b is ccorrect. For the first one are you allowed to use the mean value theorem? If so then you might try aplying it to this problem.

3. Nov 10, 2007

### 2RIP

Well the Mean Value Theorem does not mention anything about f(x)=f(y)... so that's why I'm not sure. Is there some other theorem out there that can justify this?

Edit: Oh nevermind, I believe it's Rolle's theorem.

4. Nov 10, 2007

### d_leet

What does the mean value theorem say?

5. Nov 10, 2007

### 2RIP

Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one point c in (a, b) where f '(c) = 0.

The question says, "If f is defined," does that necessarily mean that f is continuous?

6. Nov 10, 2007

### 2RIP

Oops, Sorry that was not MVT. MVT states that if f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) then there is a number c in (a,b) such that f(b)−f(a)=f′(c)(b−a).

7. Nov 10, 2007

### d_leet

No, but f is differentiable and differetiability implies continuity.

8. Nov 10, 2007

### brh2113

Part (a) states Rolle's Theorem exactly. So, if you just want to know if this is true, then the answer is yes. Or do you have to prove it? What exactly are you trying to find out?