# Positive derivative and strictly increasing

1. Aug 30, 2010

### AxiomOfChoice

Isn't it true that if a function $f(t)$ satisfies $f'(t) > 0$ on some interval, then $f$ is STRICTLY increasing on that interval; i.e., that $s < t$ implies that $f(s) < f(t)$?

2. Aug 30, 2010

### Office_Shredder

Staff Emeritus
You can prove this using the mean value theorem, by assuming there are points s and t with $$s<t$$ and $$f(s)\geq f(t)$$ and developing a contradiction

3. Aug 30, 2010

### StalkerM

Yes.
In fact:
if x2>x1 implies that
f(x2)>f(x1)

the derivative is:
[f(x2)-f(x1)]/(x2-x1)
(as x2 approach x1)

The numerator is positive and also the denominator (by hypothesis ) implies that:
f'(x)>0

4. Aug 30, 2010

### AxiomOfChoice

Thanks, guys. I was pretty sure this was a MVT exercise, but I was so tired last night I couldn't force myself to get out of bed and look for paper and pencil to run through it :) I'm amazed I was actually able to phrase the question correctly.

5. Jan 16, 2011

### Ryker

Related to the topic at hand, if, on the other hand, f'(a) > 0, is there necessarily an interval of positive length that is centered at c, on which f is strictly increasing? I've though about this for a while now, and it seems to me the answer is no, but I just can't come up with a counterexample. I can imagine an ever oscillating function that never reaches f(a) when x > a, but since it constantly oscillates, you also can't pinpoint the endpoint of an interval, on which f would be strictly increasing.

Any thoughts on this, is this a good counterexample? And does anyone perhaps have a specific function in mind, one that isn't just vaguely described such as above in my previous paragraph?

Last edited: Jan 16, 2011