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Maxima and Minima
f has a relative maximum at c if there is some interval (r, s) (even a very small one) containing c for which f(c) |
Sometimes it is not a simple matter to tell from the graph Exactly where the local extrema are situated. For instance, try graphing the curve y = x3(x1/2 -1) for 0 x
2, and see if you can tell exactly where the absolute minimum lies. (This one is discussed in the on-line review exercises).
To help us locate extrema accurately, we classify them into three types and use calculus to assist us in locating them.
Locating Candidates for Relative Extrema
If f is continuous on its domain and differentiable except at a few isolated points, then its relative extrema occur among the following types of points.
1. Stationary points: points x in the domain where f'(x) = 0. To locate stationary points, set f'(x) = 0 and solve for x.
The following figure shows several instances of all three types. ![]() |
Example 2 in the book goes applies this to the function f(x) = x4 - 4x3 we looked at above. Instead of repeating that example, let us go through a different example.
Stationary Points Set the derivative f'(x) = 0 and solve for x.
You should get three solutions (enter them in any order).
Now classify these three stationary points as relative maxima, minima, or neither. (Make all 6 selections and press "Check".)
Singular points To locate singular points, we look for values x where f'(x) is not defined, but f(x) is defined. However, f'(x) = 4x - 4x3 is defined for every x in the domain. Thus, there are no singular points.
Endpoints These are the endpoints of the domain, if any. Since the domain is [-1, +), it has only one end-point: x = -1.
Q Looking at the graph, we see that, at x = -1, f has a
Q OK what about those mysterious singular points? I haven't seen any yet.
A Try the following example.
Notice that something funny seems to be going on around x = 1. (We will zoom in later, after we have done the calculations.)
Try to identify the location of the statinoary point(s) on the graph before going further.
Singular points To locate singular points, we look for values x where f'(x) is not defined, but f(x) is defined. Now, you should laready have calculated f'(x) to answer the last question. It is
f'(x) = | 3(x-1)1/3 | - 3 |
Notice that the denominator is zero when x = 1, so that f'(x) is not defined when x = 1, even though f(x) is defined when x = 1 (You should calculate f(1) = 0). Thus, we have a singular point at x = 1.
Endpoints The only endpoint is x = 0.
Here is a zoomed-in portion of the graph.
Notice that we have a relative minimum at the singular point x = 1, and a relative maximum at the stationary point a little to its right.