Differentiable on interval implies monotonic on some neighborhood of every point

[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..

 

[jedishrfu]

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?

 

[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).

 

[jedishrfu]

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.

 

[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.

 

[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.

 

[JG89]

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

 

[Poopsilon]

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

 

[lavinia]

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