Differentiable on interval implies monotonic on some neighborhood of every point

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Discussion Overview

The discussion centers on the implications of differentiability and the positivity of the derivative at a point within an interval. Participants explore whether differentiability on an interval and a positive derivative at a point imply that the function is monotonically increasing in some neighborhood of that point. The scope includes theoretical reasoning and mathematical exploration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that if \( f \) is differentiable on \([a,b]\) and \( f'(c) > 0 \) for some \( a < c < b \), then \( f \) should be monotonically increasing in some neighborhood of \( c \), though they struggle to prove it.
  • Another participant presents the sine function as a counterexample, noting that while it is differentiable and has a positive derivative on \((0, \pi/2)\), it is not monotonically increasing over the entire interval \((0, \pi)\).
  • Some participants agree that there exists a neighborhood around points where \( f'(x) > 0 \) within \((0, \pi/2)\) where the sine function is indeed monotonically increasing.
  • One participant argues that while there may be local monotonicity at \( c \), it does not imply that there is a neighborhood where \( x < y \) leads to \( f(x) < f(y) \) for all \( x, y \) in that neighborhood.
  • Another participant suggests that if the derivative is continuous at \( c \), then the implication of monotonicity in a neighborhood holds true.

Areas of Agreement / Disagreement

Participants express differing views on whether differentiability and a positive derivative imply local monotonicity. Some agree that there can be neighborhoods of monotonicity, while others challenge this notion, indicating that the discussion remains unresolved.

Contextual Notes

Participants reference specific functions and their behaviors, such as the sine function and proposed examples like \( x^2 \sin(1/x) \), to illustrate their points. There is an acknowledgment of the need for conditions like continuity of the derivative to support claims about monotonicity.

Poopsilon
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If f is differentiable on [a,b] and f'(c)>0 for some a<c<b then does this imply that f is monotonically increasing on some neighborhood of c? My intuition says yes but I just can't figure out a way to prove it. (not homework). Because of the weierstrass function I'm pretty sure differentiability on the whole neighborhood has to be utilized in some way..
 
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what about a sin curve from 0 to pi, its differentiable over the interval and sin' x > 0 when 0<x<pi/2 and
yet its not monotonically increasing from 0 to pi but there is a neighborhood about c where it is?
 
I'm not sure what you're trying to say here. Your correct, sin(x) is not monotonically increasing from 0 to pi, but at any point in the interval (0,pi/2) (which are the only points from 0 to pi where sin'(x) is positive) there is a neighborhood around that point on which sin(x) is monotonically increasing, trivial in this case since sin(x) is monotonically increasing on all of (0,pi/2).
 
Poopsilon said:
I'm not sure what you're trying to say here. Your correct, sin(x) is not monotonically increasing from 0 to pi, but at any point in the interval (0,pi/2) (which are the only points from 0 to pi where sin'(x) is positive) there is a neighborhood around that point on which sin(x) is monotonically increasing, trivial in this case since sin(x) is monotonically increasing on all of (0,pi/2).

Yeah I think you're right I was trying to find a counterexample and sin seemed to fit but you're right no it doesn't.
 
Yes. If f'(c) > 0 then [itex]\frac{f(c+h) - f(c)}{h} > 0[/itex] on some neighborhood of c. If h > 0 then [itex]\frac{f(c+h) - f(c)}{h} > 0 \Rightarrow f(c+h) - f(c) > 0 \Rightarrow f(c+h) > f(c)[/itex] where c < c +h. If h < 0, then [itex]\frac{f(c+h) - f(c)}{h} > 0 \Rightarrow f(c+h) - f(c) < 0 \Rightarrow f(c+h) < f(c)[/itex] where c + h < c. This implies monotonicity on the given neighborhood of c.
 
i don't think so. it is locally monotonic "at c" in the sense that f(x) < f(c) for x<c and close to c, and same for greater, but there is no nbhd on which f is monotonic.

I.e. my intuition says there is no nbhd in which x<y implies f(x) < f(y). I would try to take smth

like x^2.sin(1/x) and tilt it a little.
 
Sorry OP, I made my statement too quick. What I proved does not imply monotonicity.
 
Yah you're right mathwonk, tilt it a bit and set f(x)=0 for x=0.
 
Poopsilon said:
If f is differentiable on [a,b] and f'(c)>0 for some a<c<b then does this imply that f is monotonically increasing on some neighborhood of c? My intuition says yes but I just can't figure out a way to prove it. (not homework). Because of the weierstrass function I'm pretty sure differentiability on the whole neighborhood has to be utilized in some way..

If the derivative is continuous at c then it works because then it is positive in an interval around c
 

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