Prove existence of open interval with non-zero derivative

In summary, the conversation revolves around a proof involving an open interval D containing x0 where f'(x) ≠ 0 for all x∈D, given f'(x0)≠0. The conversation includes examples, counter-examples, and discussions about the limit definition of the derivative. It is suggested that there may be a restriction on functions for which the statement can be proved true.
  • #1
meriadoc
5
0
I'm struggling to get started with the proof that an open interval D containing x0 exists such that f'(x) ≠ 0 for all x∈D, given f'(x0)≠0. It seems like it should be easy but I've been scratching around for an hour now and have gotten nowhere, could anyone give me some advice to help me along?
 
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  • #2
meriadoc said:
I'm struggling to get started with the proof that an open interval D containing x0 exists such that f'(x) ≠ 0 for all x∈D, given f'(x0)≠0. It seems like it should be easy but I've been scratching around for an hour now and have gotten nowhere, could anyone give me some advice to help me along?
I can't believe that this is the correct statement of the problem. Consider the open interval (0, 2), with x0 = 1, and f(x) = x. For all x in this interval, f'(x) ≠ 0, and f(x0) ≠ 0. Are you sure you wrote the problem correctly?

BTW, if this is a homework problem it should have been posted in the Homework & Coursework section, not here in the technical math sections.
 
  • #3
While the derivative of an arbitrary differentiable function is not necessarily continuous, it can be proved that it has the "intermediate value property". That is, if f(a)= c and f(b)= d, then f takes on every value between c and d at some point between a and b.

Mark44, did you misread the problem? What you give is an example for which the "theorem" works, not a counter-example.
 
  • #4
Assume the opposite. Would f be differentiable at x0?
 
  • #5
HallsofIvy said:
While the derivative of an arbitrary differentiable function is not necessarily continuous, it can be proved that it has the "intermediate value property". That is, if f(a)= c and f(b)= d, then f takes on every value between c and d at some point between a and b.

Mark44, did you misread the problem? What you give is an example for which the "theorem" works, not a counter-example.
It wasn't clear to me what the OP was trying to prove (quoted below), and was trying to get some clarification.
meriadoc said:
I'm struggling to get started with the proof that an open interval D containing x0 exists such that f'(x) ≠ 0 for all x∈D, given f'(x0)≠0. It seems like it should be easy but I've been scratching around for an hour now and have gotten nowhere, could anyone give me some advice to help me along?
 
  • #6
As I understand it, we are given an arbitrary function f. If f'(x0) is non-zero we are asked to prove that there is some open interval containing x0 throughout which f'(x) is non-zero. One might assume that f is everywhere differentiable.

A function f such that f(0) = 0, f'(0) = 1 and f'(x) = 0 for an infinite number of x's arbitrarily near zero would serve as a counter-example. Can you find a function like that?
 
  • #7
This statement is false. Take [itex]x_0=0[/itex] Consider [itex] f(x)=x^2\sin(1/x)+x[/itex] for nonzero [itex]x[/itex] with [itex] f(0)=0[/itex]. Computation gives [itex] f'(0)=1[/itex] and [itex] f'(x)=2x\sin(1/x)-\cos(1/x)+1[/itex] for nonzero [itex]x[/itex]. Observing that [itex] f'(\frac{1}{2n\pi})=0[/itex] for integer [itex]n[/itex] shows that this [itex]f[/itex] is a counterexample since any open subset of [itex]\mathbb{R}[/itex] containing [itex]0[/itex] contains elements of the form [itex]\frac{1}{2n\pi}[/itex].
 
  • #8
Infrared said:
Computation gives [itex] f'(0)=1[/itex]
Can you show that from the limit definition of the derivative?
 
  • #9
MrAnchovy said:
Can you show that from the limit definition of the derivative?

As you wish:

[tex]f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{h^2\sin(1/h)+h}{h}=\lim_{h\to 0} h\sin(1/h)+1=1[/tex]

since [itex]\lim_{h\to 0} h\sin(1/h)=0[/itex] follows from the squeeze theorem.
 
  • #10
Infrared said:
... since [itex]\lim_{h\to 0} h\sin(1/h)=0[/itex] follows from the squeeze theorem.
Yes indeed, so the original statement is in general false.

I notice the OP hasn't returned; @meridoc if you are still interested does the counterexample above suggest a restriction on functions for which the statement can be proved true?
 

1. What is an open interval?

An open interval is a set of real numbers that includes all numbers between two given numbers, but does not include the endpoints.

2. What does it mean for a function to have a non-zero derivative?

A function with a non-zero derivative means that the rate of change of the function is not equal to zero at any point. In other words, the function is not constant and is continuously changing.

3. Why is it important to prove the existence of an open interval with a non-zero derivative?

Proving the existence of an open interval with a non-zero derivative is important because it helps us understand the behavior of a function. It also allows us to determine whether a function is increasing or decreasing on a given interval.

4. How can you prove the existence of an open interval with a non-zero derivative?

To prove the existence of an open interval with a non-zero derivative, you can use the Mean Value Theorem or the Intermediate Value Theorem. These theorems provide conditions for the existence of a point in the interval where the derivative is non-zero.

5. Can you give an example of a function with an open interval and a non-zero derivative?

One example of a function with an open interval and a non-zero derivative is f(x) = x^2 on the interval (0, 3). This function has a derivative of 2x, which is non-zero for all values of x in the open interval. Therefore, this function satisfies the condition of an open interval with a non-zero derivative.

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