Increasing or Decreasing Nature of a Function

Transcription

1 Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 1/ 46 Increasing or Decreasing Nature of a Function Examining the graphical behavior of functions is a basic part of mathematics and has applications to many areas of study. When we sketch a curve, just plotting points may not give enough information about its shape. For example, the points (-1,0), (0,-1), and (1,0) satisfy the equation given by y = (x + 1) 3 (x 1). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 2/ 46

2 On the basis of these points, we might hastily conclude that the graph should appear as in Figure 1 (a), but in fact the true shape is given in Figure 1(b). Figure 1: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 3/ 46 In this chapter we will explore the powerful role that differentiation plays in analyzing a function so that we can determine the true shape and behavior of its graph. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 4/ 46

3 We begin by analyzing the graph of the function y = f(x) in Figure 2. Notice that as x increases (goes from left to right) on the interval I 1, between a and b, the values of f(x) increase and the curve is rising. Figure 2: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 5/ 46 Definition A functionf is said to be increasing on an interval I when, for any two numbers x 1, x 2 in I if x 1 < x 2, then f(x 1 ) < f(x 2 ). A function f is decreasing on an interval I when, for any two numbers x 1, x 2 in I, if x 1 < x 2, then f(x 1 ) > f(x 2 ). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 6/ 46

4 Turning again to Figure 2, we note that over the interval I 1, tangent lines to the curve have positive slopes, so f (x) must be positive for all x in I 1. A positive derivative implies thai the curve is rising. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 7/ 46 Over the interval I 2, the tangent lines have negative slopes, so f (x) < 0 for all x in I 2. The curve is falling where the derivative is negative. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 8/ 46

5 We thus have the following rule, which allows us to use the derivative to determine when a function is increasing or decreasing: Rule ( Criteria for increasing or Decreasing Function) Let f be differentiable on the interval (a, b). If then f is increasing on (a, b). f (x) > 0 for all x (a, b) If f (x) < 0 for all x (a, b) then f is decreasing on (a, b). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 9/ 46 To illustrate these ideas, we will use Rule to find the intervals on which y = 18x 2 3 x3 is increasing and the intervals on which y is decreasing. Letting y we must determine when f (x) is positive and when f (x) is negative. We have f (x) = 18 2x 2 = 2(9 x 2 ) = 2(3 + x)(3 x) Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 10/ 46

6 We can find the sign of f(x) by testing the intervals determined by the roots of 2(3 + x)(3 x) = 0, namely, 3 and 3. These should be Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 11/ 46 Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 12/ 46

7 These results are indicated in the sign chart, where the bottom line is a schematic version of what the signs of f say about f itself. Notice that the horizontal line segments in the bottom row indicate horizontal tangents for f at 3 and at 3. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 13/ 46 Thus, f is decreasing on (, 3) and (3, ) and is increasing on ( 3, 3). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 14/ 46

8 This corresponds to the rising and falling nature of the graph of f shown in Figure 3. Figure 3: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 15/ 46 Example Example: Find where the function f(x) = 3x 4 4x 3 12x is increasing and where it is decreasing.. Solution: f (x) = 12x 3 12x 2 24x = 12x(x 2)(x + 1) To use the I/D Test we have to know where f (x) > 0 and where f (x) < 0. This depends on the signs of the three factors of f (x), namely, 12x, x 2 and x + 1. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 16/ 46

9 Example... We divide the real line into intervals whose endpoints are the critical numbers 1, 0, 2 and arrange our work in a chart. Decreasing Increasing Decreasing Increasing A plus sign indicates that the given expression is positive, and a minus sign indicates that it is negative. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 17/ 46 Example... Therefore, the function f(x) = 3x 4 4x 3 12x is DECREASING on (, 1), is INCREASING on ( 1, 0), is DECREASING on (0, 2), is INCREASING on (2, ). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 18/ 46

10 Look now at the graph of y = f(x) in Figure 4. Some observations can be made. First, there is something special about the points P, Q, and R. Figure 4: Notice that P is higher than any other nearby point on the curve and likewise for R. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 19/ 46 The point Q is lower than any other nearby point on the curve. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 20/ 46

11 Since P, Q, and R may not necessarily be the highest or lowest points on the entire curve, we say that the graph of f has relative maxima at a and at c, and has a relative minimum at b. The function f has relative maximum values of f(a) at a and f(c) at c; and has a relative minimum value of f(b) at b. We also say that (a, f(a)) and (c, f(c)) are relative maximum points and (b, f(b)) is a relative minimum point on the graph off. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 21/ 46 Turning back to the graph, we see that there is an absolute maximum (highest point on the entire curve) at a, but there is no absolute minimum (lowest point on the entire curve) because the curve is assumed to extend downward indefinitely. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 22/ 46

12 Definition A function f has a relative maximum at a if there is an open interval containing a on which f(a) > f(x) for all x in the interval. The relative maximum value is f(a). A function f has a relative minimum at a if there is an open interval containing a on which f(a) < f(x) for all x in the interval. The relative minimum value is f(a). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 23/ 46 Definition A function f has an absolute maximum at a if f(a) > f(x) for all x in the domain of f. The absolute maximum value is f(a). A function f has an absolute minimum at a, if f(a) < f(x) for all x in the domain of f. The absolute minimum value is f(a). Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 24/ 46

13 We refer to either a relative maximum or a relative minimum as a relative extremum (plural: relative extrema). Similarly, we speak of absolute extrema. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 25/ 46 Referring to Figure 4, we notice that at a relative extremum the derivative may not be defined (as when x = c). But whenever it is defined at a relative extremum, it is 0 (as when x = a and when x = b) and hence the tangent line is horizontal. We can state the following: Rule (A Necessary Condition for Relative Extrema) If f has a relative extremum at a, then f (a) = 0 or f (a) does not exist. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 26/ 46

14 Rule does not say that if f(a) is 0 or f(a) does not exist, then there must be a relative extremum at a. In fact, there may not be one at all. For example, in Figure 5(a), f (a) is 0 because the tangent line is horizzontal at a, but there is no relative extremum there. Figure 5: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 27/ 46 In Figure 5(b), f (a) does not exist because the tangent line is vertical at a, but again there is no relative extremum there. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 28/ 46

15 Definition For a in the domain of f, if either f (a) = 0 or f(a) does not exist, then a is called a critical value for f. If a is a critical value, then the point (a, f(a)) is called a critical point for f. At a critical point, there may be a relative maximum, a relative minimum, or neither. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 29/ 46 Rule Suppose f is continuous on an open interval I that contains the critical value a and f is differentiable on I, except possibly at a. 1 If f (x) changes from positive to negative as x increases through a, then f has a relative maximum at a. 2 If f (x) changes from negative to positive as x increases through a, then f has a relative minimum at a. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 30/ 46

16 Example Test y = f(x) = x 2 e x for relative extrema. By the product rule f (x) = e x (2x) + x 2 e x = xe x (x + 2) Noting that e x is always positive, we obtain the critical values 0 and 2. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 31/ 46 From the sign chart of f(x) given in Figure 6, we conclude that there is a relative maximum when x = 2 and a relative minimum when x = 0. Figure 6: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 32/ 46

17 Example Sketch the graph of y = f(x) = 2x 2 x 4 with the aid of intercepts, symmetry, and the first-derivative test. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 33/ 46 Example Test y = F (x) = x 2/3 for relative extrema. We have f (x) = 2 3 x 1/3 = x Since f (0) does not exist, x = 0 is critical point. And there is no real number such that f (x) = 0. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 34/ 46

18 So there is only one critical point. The sign chart is given in the following figure: Figure 7: Since f(x) is defined at x = 0, f has a relative minimum at 0 of f(0) = 0, and there are no other relative extrema. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 35/ 46 Example If c = 3q 3q 2 + q 3 is a cost function, when is marginal cost increasing? Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 36/ 46

19 Example (Storage and Shipping Costs) In his model for storage and shipping costs of materials for manufacturing process, Lancaster derives the cost function ( C(k) = k ) 1 k 100 k Where C(k) is the total cost (indollars) of storage and transportation for 100 days of operation if a load of k tons of material is moved every k days. 1 Find C(1) 2 For what values of k does C(k) have a minimum? 3 What is the minimum value? Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 37/ 46 Absolute Extrema on a Closed Interval Theorem (Extreme-Value Theorem) If a function is continuous on a closed interval, then the function has both a maximum value and a minimum value on that interval. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 38/ 46

20 For example, each function in Figure 8 is continuous on the closed interval [1, 3]. Geometrically, the extreme-value theorem assures us that over this interval each graph has a highest point and a lowest point. Figure 8: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 39/ 46 We will focus our attention on absolute extrema and make use of the extreme-value theorem where possible. If the domain of a function is a closed interval, to determine absolute extrema we must examine the function not only at critical values, but also at the endpoints. Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 40/ 46

21 For example. Figure 9 shows the graph of the continuous function y = f(x) over [a, b]. The extreme-value theorem guarantees absolute extrema over the interval. Clearly, the important points on the graph occur at x = a, b, c, and d, which correspond to endpoints or critical values. Figure 9: Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 41/ 46 Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 42/ 46

22 Example Find absolute extrema for f(x) = x 2 4x + 5 over the closed interval [1, 4] To find the critical values of f, we must find f f (x) = 2x 4 = 2(x 2) this gives the critical value x = 2. Evaluating f(x) at the end points 1 and 4 and at the critical value 2, we have f(1) = 2 f(4) = 5 f(2) = 1 From the values, we conclude that the maximum is f(4) = 5 and the minimum is f(2) = 1 Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 43/ 46 Example Find absolute extrema for f(x) = 3x 5 + 5x 3 over the closed interval [ 2, 0] Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 44/ 46

23 Example (Maximizing Revenue) The demand equation for a manufacturer s product p = 80 q 4 0 q 80 where q is the number of units and p is the price per unit. At which value of q will there be maximum revenue? What is the maximum revenue? Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 45/ 46 Example Suppose that the demand equation for a monopolist s product is p = 400 2q and the average-cost function is c = 0.2q (400/q). where q is number of units, and both p and c are expressed in dollars per unit. 1 Determine the level of output at which profit is maximized. 2 Determine the price at which maximum profit occurs. 3 Determine the maximum profit. 4 If, as a regulatory device, the government imposes a tax of $22 per unit on the monopolist, what is the new price for profit maximization? Öğr. Gör. Volkan ÖĞER FBA 1021 Calculus 46/ 46