# 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