Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiable on interval implies monotonic on some neighborhood of every point

  1. Jan 14, 2012 #1
    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..
     
  2. jcsd
  3. Jan 14, 2012 #2

    jedishrfu

    Staff: Mentor

    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?
     
  4. Jan 14, 2012 #3
    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).
     
  5. Jan 14, 2012 #4

    jedishrfu

    Staff: Mentor

    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.
     
  6. Jan 14, 2012 #5
    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.
     
  7. Jan 14, 2012 #6

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  8. Jan 14, 2012 #7
    Sorry OP, I made my statement too quick. What I proved does not imply monotonicity.
     
  9. Jan 14, 2012 #8
    Yah you're right mathwonk, tilt it a bit and set f(x)=0 for x=0.
     
  10. Jan 14, 2012 #9

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    If the derivative is continuous at c then it works because then it is positive in an interval around c
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differentiable on interval implies monotonic on some neighborhood of every point
Loading...