Fundamental Theorem of Calculus properties

Click For Summary

Homework Help Overview

The discussion revolves around finding a function defined on the interval [-1,1] that is continuous, differentiable on (-1,1), but has a derivative that is not differentiable on that same interval. This relates to properties of the Fundamental Theorem of Calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the Mean Value Theorem and the properties of the function in question. There are attempts to connect the requirements of the problem with the Fundamental Theorem of Calculus. Some suggest starting with a nondifferentiable function and integrating it to find the desired function.

Discussion Status

Participants are actively discussing potential functions that meet the criteria, with some suggesting the absolute value function as a candidate. There is acknowledgment of the need for a function that is once differentiable but not twice differentiable, indicating a productive exploration of ideas.

Contextual Notes

There is a focus on the properties of differentiability and continuity, with specific attention to the behavior of the function at the point x=0. Participants are navigating the constraints of the problem without reaching a consensus on a specific solution.

am100181
Messages
3
Reaction score
0

Homework Statement



Find a function f : [-1,1] ---> R such that f satisfies the following properties:

a) f is continuous
b) f is restricted to (-1,1) is differentiable
c) its derivative f' is not differentiable on (-1,1)

Homework Equations




The Attempt at a Solution


I kinda think that the mean value theorem and Theorem 2 of the fundamentals \intf(x)dx = F(b)-F(a) got some link but I can't seem to get it. I do understand that for f'' not to exist, x should be undefined on the (-1,1). Please help.
 
Physics news on Phys.org
A read somewhere that a hint would be to begin with an absolute value and use \intf(x)dx = F(b)-F(a) (fundamental theorem of calc prep 2) repeatedly.. but still puzzled
 
Start with c). Pick a nondifferentiable function on (-1,1) and integrate it to get f.
 
|x| is a very simple function that is not differentiable at x= 0.
 
HallsofIvy said:
|x| is a very simple function that is not differentiable at x= 0.

You are right, but the OP is looking for a function such that it is once differentiable on (-1,1) but not twice, and is continuous of course on the same interval.

Edit: ignore it!
 
Yes, and combining |x| with Dick's suggestion gives exactly that!

(Edit: Too late! I gotcha!)
 

Similar threads

Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K