Absolute and Local Extrema

Transcription

1 Extrema of Functions We can use the tools of calculus to help us understand and describe the shapes of curves. Here is some of the data that derivatives f (x) and f (x) can provide about the shape of the curve f(x): 1. highest and lowest points 2. where f(x) is increasing or decreasing 3. where f(x) is concave up or concave down In this chapter, we will learn how to use f and f to gather this data; we can put this data together with information that we will find about asymptotes of f(x) to draw the curve defined by f(x) with a great deal of accuracy. Absolute and Local Extrema Consider the graph of the function f(x) = x 3 /3 x, 3 x 3: There are several interesting things to note about this graph. First of all, f(x) has a point that is higher than any other on its graph this point has coordinates x = 3, f(3) = 6. Notice that f(x) also has a point that is lower than any other point on its graph; this point has coordinates x = 3 and f( 3) = 6. The point on the graph of f(x) at x = 1 (where f(1) =.6) is also interesting; while it is not the lowest point on the curve, it is the lowest point on the curve near x = 1. 1

2 Similarly, the point on the graph of f(x) at x = 1 (when f( 1) =.6) is not the highest point on the curve, but we see that this point is the highest point on the curve near x = 1. All of these types of points are important, so we will give them names: Definition 1. Let c be in the domain of f(x). If f(c) f(x) for every number x in the domain of f, then f(c) is the absolute maximum value of f(x). If f(c) f(x) for every number x in the domain of f, then f(c) is the absolute minimum value of f(x). Definition 2. Let c be in the domain of f(x). If f(c) f(x) for every number x near x = c, then f(c) is a local maximum value of f(x). If f(c) f(x) for every number x near x = c, then f(c) is a local minimum value of f(x). In particular, every absolute extreme is also a local extreme. Let s use these definitions to classify the points on the graph of f(x) = x 3 /3 3 that we looked at earlier on the interval 3 x 3; the function is graphed below: 2

3 f(x) has absolute maximum value 6 f(x) has local minimum value 6.6 is a local maximum value of f(x).6 is a local minimum value of f(x). We would like to have a method for finding a function s absolute and local extremes. However, there are some intricacies here; to understand them, consider the graph of f(x) = tan x below, π/2 < x < π/2: 3

4 Notice that, on the open interval ( π/2, π/2), f(x) has no absolute extremes (or local extremes); for example, near π/2, the function values increase towards infinity, so there s no one highest point on the curve. Now think about what happens if we look at the same function f(x) = tan x on some closed interval, say [ π/4, π/4]: Notice that, on this closed interval, f(x) now has both an absolute extreme value, f(π/4) = 1, and an absolute minimum value, f(π/2) = 1. This illustrates the concept of the Extreme Value Theorem: Theorem 3, Extreme Value Theorem. If f(x) is continuous on a closed interval [a, b], then f(x) has both an absolute maximum value f(c) and an absolute minimum value f(d), where c and d are numbers in the interval [a, b]. In other words, if our function is continuous, and the interval of interest is closed (i.e. includes its endpoints), then we can be certain that we will be able to locate an absolute max and an absolute min on the interval. The goal of the remainder of this section is to determine how to locate the absolute extremes of a continuous function on a closed interval. There are two types of points at which a function can have an absolute extreme on a closed interval. To understand this, let s think about the function f(x) = 3 sin x on two different intervals. We ll start with the interval [ π, π], graphed below: 4

5 Notice that the highest point of f(x) on this interval occurs at x = π/2, where f(x) has a local maximum, and the lowest point occurs at x = π/2, where f(x) has a local minimum: So the absolute minimum value of f(x) on [ π, π] is f( π/2) = 3, and the absolute maximum value of f(x) on the interval if f(π/2) = 3. Now let s think about the same function on the interval [ π/4, π/4]; the function is graphed below: Notice that changing the interval changed the absolute extremes; then now occur at the endpoints x = π/4 and x = π/4: 5

6 So the absolute maximum of f(x) on [ π/4, π/4] is f(π/4) = 3 2/2, and the absolute minimum of f(x) on the same interval is f( π/4) = 3 2/2. The examples above illustrate the following point: Finding Absolute Extremes of a Continuous Function on a Closed Interval The absolute extremes of a function f(x) that is continuous on the interval [a, b] must occur: either at the endpoints x = a or x = b, or at x values at which f(x) has local extremes. So in order to be able to find the absolute extremes of a continuous function f(x) on a closed interval, we must be able to locate the function s local extremes. In the remainder of this section, we will determine how to do this, then learn a method for finding the absolute extremes of f(x). Local Extremes We would like to be able to use calculus to find local maximum and local minimum values of f(x). To do so, let s think about the graph of f(x) = x 3 /3 x again: 6

7 We saw that f(x) had a local maximum value when x = 1, and a local minimum when x = 1. Notice what happens when we draw tangent lines to f(x) at x = 1 and x = 1: Both of the tangent lines are horizontal, i.e. the slope of both lines is 0. But since f (x) tells us the slope of tangent lines to f(x), we already know that f ( 1) = 0 and f (1) = 0. At this point, we might guess that finding all numbers a so that f (a) = 0 will give us all of the local extremes of f(x); however, the following example shows us that this is not the case. Consider the function f(x) = x, graphed below on the interval [ 1, 1]. 7

8 The function clearly has a local minimum at x = 0, but f(x) is not differentiable at x = 0; i.e. f (0) DNE. From this example, we see that locating local extremes of a function is slightly more complicated than just finding the numbers c so that f (c) = 0; we will definitely have to consider the numbers c so that f (c) DNE. We may be worried that there are other ways in which a function can have a local extreme; fortunately, Fermat s Theorem rules out this possibility: Theorem 4, Fermat s Theorem. If f(x) has a local maximum or local minimum at x = c, and if f (c) exists, then f (c) = 0. Fermat s Theorem tells us that the local extremes of f(x) can only occur at: 1. points c so that f (c) = 0, and 2. points c so that f (c) DNE. The numbers c above are important enough that we will give them a name: Definition 6. A critical number of the function f(x) is any number c in the domain of f(x) so that 1. f (c) = 0, or 2. f (c) DNE. Fermat s theorem tells us that the local extremes of a function can only occur at critical points. Example. Find all critical numbers of the function f(x) = x To find the critical numbers, we need to find f (x), then find all points c so that: f (c) = 0, or f (c) DNE. 8

9 Since f (x) = 3x 2 is defined for all real numbers, we do not have to worry about the second type of critical number; the only possible critical numbers are those c so that f (c) = 0. So we must determine when 3x 2 = 0; by dividing through by 3, we see that this is the same as determining when x 2 = 0. Clearly x = 0 is the only critical number of f(x). Thus Fermat s theorem says that the only possible local extreme of f(x) = x occurs at x = 0. Let s look at a graph of the function: Notice that, while f (0) = 0, f(x) does not have a local extreme at x = 0. This illustrates an important point: A function f can only have local extremes at critical points, but not every critical point yields a local extreme of f. Absolute Extremes Earlier, we saw that, if f(x) is continuous on a closed interval [a, b], then f(x) has an absolute maximum value and an absolute minimum value on [a, b], the absolute extremes can only occur at two types of points: the endpoints a and b local extremes of f(x), which can only occur at critical numbers, i.e.: (a) numbers c so that f (c) = 0, or (b) numbers c so that f (c) DNE. Putting all of this information together, we get the following method for finding absolute extremes: 9

10 The Closed Interval Method: To find the absolute maximum and minimum values of a continuous function f(x) on a closed interval [a, b]: 1. Find the critical numbers of f(x) that occur in (a, b) 2. Evaluate f(x) at each critical number 3. Evaluate f(x) at each endpoint a and b 4. The absolute maximum of f(x) on the interval is the largest value from steps 2 and 3; the absolute minimum of f(x) on the interval is the smallest of these values. Example. Find the absolute maximum and absolute minimum values of f(x) = x2 3x + 1 x on the interval [1/3, 5]. Since the only value at which f(x) is not continuous is x = 0, which is not in [1/3, 5], we know that f has absolute extremes on the interval. We must start by finding the critical numbers of f, which means that we need to calculate f (x). Since f(x) = x2 3x + 1 x is a quotient, we can use the quotient rule to find f (x): f (x) = (2x 3)(x) (x2 3x + 1)(1) x 2 = 2x2 3x (x 2 3x + 1) x 2 = 2x2 3x x 2 + 3x 1 x 2 = x2 1 x 2. The critical numbers of f occur at number in the domain of f(x) where: 1. f (x) = 0, or 2. f (x) DNE. The only number at which f (x) does not exist is x = 0, which is not in the domain of f(x); so there are no type 2 critical numbers. To find the type 1 critical numbers, we need to determine where f (x) = 0; since f (x) is a fraction this can only occur if the numerator of f (x) is 0. So we must find all x so that x 2 1 = 0. Clearly, x = 1 and x = 1 are the only possibilities. 10

11 However, since x = 1 is not in the interval [1/3, 5], the only critical number of interest to us is x = 1. Now that we have the only the critical number, we must evaluate f(x) at this number, as well as at the endpoints of the interval. We know that: f(1) = 1 f(1/3) = 1/3 f(5) = 11/5. The largest value above is 11/5, and the smallest is 1. Thus the absolute maximum value of f(x) on the interval [1/3, 5] is 11/5, and the absolute minimum value is 1. The graph below of f(x) confirms our answers: 11