 Site Navigation                          Introductory Calculus: Information from the First Derivative Regions of Increase and Decrease AND Maximum and Minimum Values of a Function

Let's look at an example function and start by calculating is derivative: Notice that the derivative equals 0 when x = -1 or +1.

We will show that the derivative is positive when x < –1 and x > +1, and is negative when –1 < x < +1.

Below, we show the graph of twice. Notice that the graph has a peak and a valley where its derivative equaled zero, x = -1 and x = +1. On the second graph we have drawn in horizontal tangent lines (reminder: lines which are horizontal have a slope of 0). Some observations about these diagrams are made below the graphs.  1. appears to be “going up” until x = –1 and then again after x = +1.
2. appears to be “going downward” from x = –1 to x = +1.
3. The horizontal tangents appear to intersect the graph of at (–1, 2) and (1, –2).

These observations can be summarized as follows:

1. is increasing when x < –1 and x > +1
(notice the derivative is positive).
2. is decreasing when –1 < x < +1
(notice the derivative is negative).
3. has a relative maximum at (–1, 2) and a relative minimum at (1, –2)
(where the derivative equals 0).

Generalized Results
 A function is increasing at a point x = c if there is an interval containing c and the derivative is positive on this interval.    A function is decreasing at a point x = c if there is an interval containing c and the derivative is negative on this interval.    A function possibly has a relative maximum or minimum value at a point where the derivative is exactly 0 (or as we shall see, at a point where the derivative does not exist).    Points where the derivative is 0 or does not exist are called critical points.

We are going to use these results to make more detailed observations about functions and their graphs. We will particularly focus on profit, cost, revenue and related functions in our examples.

Let's use our profit function from test #1:

P(x) = 80x + 0.1x2 – (40x + 0.2x2 +2000) = –0.1x2 + 40x – 2000

To analyze this function for regions of increase/decrease and for possible maximum and minimum values, we will first calculate the derivative and look for critical points. This means that there is a possible max or min at the point (200, P(200)) = (200, \$2000). We can see from the graph of P(x) that there is a maximum, which is the vertex of this parabola, at that point. But how does the derivative show this? We examine the sign of the derivative in the intervals x < 200 and x > 200. Notice, as shown in the sign chart below, that the derivative of P is positive for x < 200 and negative for x > 200. This change in sign from + to – in the derivative denotes a maximum value where x = 200. The maximum value of P is \$2000. It also shows a change from P increasing to P decreasing. • When there is a change in sign from + to -, there would be a maximum value.
• When there is a change in sign from – to +, there would be a minimum value.
• If there is no change in sign, then there is neither a max nor a min.   Examples:   Find the regions of increase and decrease and any relative max or min points if .   Find the regions of increase and decrease and any relative max or min points if .

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