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) = x^{3} at x = 0. Slope 0 and horizontal tangent line there, but still x < y implies x^{3} < y^{3} so the function is strictly increasing.

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).

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.

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.

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 x^{3}, even though there is a point which the derivative is zero, the strictly increasing property still holds, like the first guy mentioned.