If f is a real valued function with f′(x) positive, f is increasing at x. Let y = f(x) and let s (the slope) = f′(x). Select ε = s/2, and determine h such that f(x+h) - f(x) over h is within ε of s. Within this range, the quotient always exceeds s/2, hence it is positive. The numerator has the same sign as h. When we go forward f increases, and when we go backward f decreases. Thus a positive derivative implies f is increasing, at least near x. A negative derivative implies f is decreasing.
If the derivative is 0 the point is called a critical point. It could be a maximum, or minimum, or neither, as shown by ±xn at the origin. Either branch could be increasing or decreasing, depending on the sign prepended to x and the parity of the exponent.
As a corollary, f′(x) must equal 0 if f attains a local minimum or a local maximum at x. A nonzero derivative means f is increasing or decreasing. If we want to find the highest or lowest point on a curve, we only need look where the derivative is 0, the critical points.
If you graph x3-3x, you'll see a hump to the left of the y axis and a dip to the right. Where exactly is the top of the hump and the bottom of the dip? Differentiate to get 3x2-3 = 0, hence x = ±1.