Differentiable on interval implies monotonic on some neighborhood of every point

1. Jan 14, 2012

Poopsilon

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. Jan 14, 2012

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?

3. Jan 14, 2012

Poopsilon

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).

4. Jan 14, 2012

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.

5. Jan 14, 2012

JG89

Yes. If f'(c) > 0 then $\frac{f(c+h) - f(c)}{h} > 0$ on some neighborhood of c. If h > 0 then $\frac{f(c+h) - f(c)}{h} > 0 \Rightarrow f(c+h) - f(c) > 0 \Rightarrow f(c+h) > f(c)$ where c < c +h. If h < 0, then $\frac{f(c+h) - f(c)}{h} > 0 \Rightarrow f(c+h) - f(c) < 0 \Rightarrow f(c+h) < f(c)$ where c + h < c. This implies monotonicity on the given neighborhood of c.

6. Jan 14, 2012

mathwonk

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.

7. Jan 14, 2012

JG89

Sorry OP, I made my statement too quick. What I proved does not imply monotonicity.

8. Jan 14, 2012

Poopsilon

Yah you're right mathwonk, tilt it a bit and set f(x)=0 for x=0.

9. Jan 14, 2012

lavinia

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