# Strictly increasing vs. increasing

1. Nov 13, 2011

### Shackleford

If the derivative of a function is bigger than or equal to zero on an interval, it is increasing.

If the derivative of a function is bigger than zero on an interval, it is strictly increasing.

How does it not imply the derivative is bigger than zero if that's what's in the definition?

2. Nov 13, 2011

### LCKurtz

You are asking about the converse of the statement I have bolded which is:

If f is strictly increasing on an interval then f'(x) > 0 on the interval.

Converses of theorems aren't always true, and this converse isn't. The problem is that f'(x) can be zero at a single point and not prevent the function from being strictly increasing. Look at the graph of f(x) = x3 at x = 0. Slope 0 and horizontal tangent line there, but still x < y implies x3 < y3 so the function is strictly increasing.

3. Nov 13, 2011

### Shackleford

This is silly. Forget the converse and logic. I have to use language first. When I read that sentence, I have to ask myself what does the term strictly increasing mean? Well, I would go to that definition.

Since the definition of increasing includes bigger than or equal to zero, what else would I conclude? Why is not valid to say this is an increasing function?

Oh, I forgot about that last part. It seems the derivative characterization is simply a general tool to determine if a function is strictly increasing, but there are some point-wise exceptions. The essential aspect of it is to determine is if x < y, then f(x) < f(y).

Last edited: Nov 13, 2011
4. Nov 13, 2011

### LCKurtz

OK. If my response is silly and you feel the need to explain to me how I should help you, you obviously don't need any more help from me.

5. Nov 13, 2011

### Staff: Mentor

The difference between increasing versus strictly increasing is the difference between <= and <.

If a function is increasing on an interval [a, b], and if x < y, then f(x) <= f(y).
If a function is strictly increasing on an interval [a, b], and if x < y, then f(x) < f(y).

The function f(x) = 3 would be considered increasing, but not strictly increasing. (Note that f'(x) = 0 on any interval.)

The function g(x) = 2x + 1 is strictly increasing. Note that g'(x) = 2 > 0 on any interval.

6. Nov 13, 2011

### Shackleford

No. I didn't mean your response is silly. I was talking about the problem. I forgot about the most important part which is the last thing you mentioned. Your response elucidated it for me. Thanks.

7. Nov 13, 2011

### Shackleford

Yeah, I was being a bit singleminded in this by just focusing on the derivative condition in the theorem and statement. I forgot that what determines strictly increasing is if x < y, then f(x) < f(y). In the case of x3, even though there is a point which the derivative is zero, the strictly increasing property still holds, like the first guy mentioned.

8. Nov 14, 2011

### LCKurtz

OK, I can go with that. Thanks for clarifying.