Ok this has been killing me for a while.
Here's a random graph to illustrate my points.
So this is f'(x). At points x=-3 and x=-2 the function is increasing. What gets me is that my teacher says it's a close function or [-3,-2].
In my opinion it's (-3,-2). Here are the definitions we use:
A function f is (strictly) increasing on an interval I if for every x1x2 e I with x1 < x2, f(x1) < f(x2) [ie., f(x) gets larger as x gets larger]
Now using this definition I can see how it's a closed interval. But we are given the derivative of the graph, so here's the Theorem for a derivative of a graph.
Suppose that f is differentiable on an interval I.
(i) if f'(x) > 0 for all x e I, then f is increasing on I.
So let's use [-3,-2] for I. At the points -3 and -2, it is not differentiable because we can't find the derivative at endpoints. So wouldn't it make this an open interval (-3,-2)? Or if we use an open interval outside to the left os -3 and to the right of -2, the derivative at -3 and -2 is zero, so the theorem these points can't be included because they are not increasing according to theorem 4.1! This has me so puzzled. Which do I use and what about theorem 4.1?