Rolle's theorem -> Differentiability

Click For Summary

Homework Help Overview

The discussion revolves around verifying the hypotheses of Rolle's theorem, specifically focusing on the differentiability of the function f(x) = √(x) - (1/3)x over the interval [0, 9]. The original poster expresses uncertainty about how to prove differentiability, despite understanding continuity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish that the function is differentiable by asserting it is defined and continuous on the interval. However, participants question this assumption, noting that continuity does not imply differentiability. They suggest examining the derivative to identify any potential issues within the interval.

Discussion Status

Participants are exploring the conditions for differentiability and discussing the implications of the derivative's formula. There is a recognition of the need to verify differentiability specifically over the interval [0, 9], with some guidance provided on considering the behavior of the derivative.

Contextual Notes

There is a focus on the specific interval [0, 9] and the need to check for any points within this interval that may affect differentiability, particularly at the endpoints or critical points of the function.

Spiralshell
Messages
15
Reaction score
0

Homework Statement


So I'm doing problems where I have to verify Rolle's hypotheses. I am only having trouble with the differentiability part. My professor wants me to prove this. So for example,
f(x)=√(x)-(1/3)x [0,9]

Homework Equations


none

The Attempt at a Solution


1.) I know the function is continuous because root functions are continuous on their domains and polynomials are continuous everywhere therefore the difference of two continuous functions is continuous.

2.) What could I say about it being differentiable?
My GUESS: it is defined everywhere on its domain [0,9] and continuous it is therefore differentiable on its domain [0,9]?

I am only stuck on 2.) so no need to go further on the problem.
 
Physics news on Phys.org
No. Saying that a function is "defined and continuous" on a domain does NOT mean that it is differentiable on the interval. For example, |x| is "defined and continuous" for all x but is not differentiable at x= 0.

It's hard to tell you how you should answer this without knowing what you about derivatives. IF you know that the derivative of x^n is nx^{n-1} then it should be easy to tell what the derivative of x^{1/2}- x/3 is and so where it is differentiable.

(It is NOT differentiable on [0, 1] but Rolle's theorem does not require it to be.)
 
I know the derivative to the function. I also know chain rule and the limit definition as well quotient rule, etc. So the way my professor explained is that I have to verify I can differentiate over the interval given. In this case [0,9].

So I know f'(x)=1/(2√x)-(1/3)

I just don't understand how I verify that is differentiable over [0,9].

Here is how he gave me Rolle's Theorem (very simply):
1.) f is continuous on [a,b] (Prove it!)
2.) f is differentiable on (a,b) (Prove it!)
3.) f(a)=f(b) then if all 3 conditions are met a<c<b such that c belongs to (a,b) and that f'(c)=0.

I hope this helps you understand my situation. I am only stuck on 2.)... The rest is I understand.

Also, thank you for replying. I appreciate all the help I can get.
 
Spiralshell said:
I know the derivative to the function. I also know chain rule and the limit definition as well quotient rule, etc. So the way my professor explained is that I have to verify I can differentiate over the interval given. In this case [0,9].

So I know f'(x)=1/(2√x)-(1/3)

I just don't understand how I verify that is differentiable over [0,9].
Well, you have computed a formula for the derivative. Are there any values of ##x## in ##[0,9]## that cause problems?
 

Similar threads

Replies
3
Views
2K
Replies
26
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K