Mean Value Theorem and Rolle's Theorem: Conditions and Examples

[2RIP]

Homework Statement
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.

 

[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.

 

[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.

 

[d_leet]

What does the mean value theorem say?

 

[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?

 

[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).

 

[d_leet]

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

No, but f is differentiable and differetiability implies continuity.

 

[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?