5.2 Relative Extrema • Find Relative Extrema of a function using the first derivative Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Relative Extrema If a function f has a relative extreme value f (c) on an open interval; then c is a critical value. So, f (c) = 0 or f (c) does not exist. Relative extrema will only occur at points where the derivative is = 0 or where it is undefined. Relative Minimum Let I be the domain of f : f (c) is a relative minimum (bottom of a valley) if there exists within I an open interval I1 containing c such that f (c) ≤ f (x) for all x in I1; f has a relative minimum at c if f (x) < 0 on (a, c) and f (x) > 0 on (c, b). That is, f is decreasing to the left of c and increasing to the right of c. If the graph is continuous (no break) at the point where the function changes from decreasing to increasing, that point is called a relative minimum point Relative Maximum Let I be the domain of f : F (c) is a relative maximum (top of a hill) if there exists within I an open interval I2 containing c such that f (c) ≥ f (x) for all x in I2. f has a relative maximum at c if f (x) > 0 on (a, c) and f (x) < 0 on (c, b). That is, f is increasing to the left of c and decreasing to the right of c. If the graph is continuous (no break) at the point where the function changes from increasing to decreasing, that point is called a relative maximum point. Using First Derivatives to Find Maximum and Minimum Values Graph over the interval (a, b) a a c c f (c) Sign of f (x) Sign of f (x) for x in (a, c) for x in (c, b) Increasing or decreasing Relative minimum – + Decreasing on (a, c]; increasing on [c, b) Relative maximum + – Increasing on (a, c]; decreasing on [c, b) b b Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 7 Using First Derivatives to Find Maximum and Minimum Values Graph over the interval (a, b) f (c) No relative maxima or minima a c c – – Increasing or decreasing Decreasing on (a, b) Increasing on (a, b) b No relative maxima or minima a Sign of f (x) Sign of f (x) for x in (a, c) for x in (c, b) + + b Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 8 Using First Derivatives to Find Maximum and Minimum Values Example 1: For the function f given by f (x) 2x 3 3x 2 12x 12. find the relative extrema. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 9 Using First Derivatives to Find Maximum and Minimum Values Example 1 (continued): Find Derivative And set it = 0 6x 2 6x 12 0 x2 x 2 (x 2)(x 1) x2 or 0 0 x 1 These two critical values partition the number line into 3 intervals: A (– ∞, –1), B (–1, 2), and C (2, ∞). A B -1 C 2 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 10 Using First Derivatives to Find Maximum and Minimum Values Example 1 (continued): 3rd analyze the sign of f (x) in each interval. Interval A C B -1 x 2 Test Value x = –2 x=0 x=4 Sign of f (x) + – + Result f is increasing f is decreasing on on (–∞, –1] [–1, 2] Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley f is increasing on [2, ∞) Slide 2.1- 11 Using First Derivatives to Find Maximum and Minimum Values Example 1 (concluded): Therefore, by the First-Derivative Test, f has a relative maximum at x = –1 given by f (1) 2(1)3 3(1)2 12(1) 12 19 The relative maximum value is 19. It occurs where x = -1. And f has a relative minimum at x = 2 given by f (2) 2(2)3 3(2)2 12(2) 12 8 The relative minimum value is -8. It occurs where x is 2. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 12 Using First Derivatives to Find Maximum and Minimum Values Example 1 (continued): Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 13 Using First Derivatives to Find Maximum and Minimum Values Example 3: Find the relative extrema for the Function f (x) given by f (x) (x 2)2 3 1 Then sketch the graph. 1st find f (x). 1 3 2 f (x) x 2 3 2 f (x) 3 3 x2 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 14 Using First Derivatives to Find Maximum and Minimum Values Example 3 (continued): 2nd find where f (x) does not exist or where f (x) = 0. Note that f (x) does not exist where the denominator equals 0. Since the denominator equals 0 when x = 2, x = 2 is a critical value. f (x) = 0 where the numerator equals 0. Since 2 ≠ 0, f (x) = 0 has no solution. Pearson Education, Thus, x = 2 is theCopyright only© 2008 critical value. Inc. Publishing as Pearson AddisonWesley Slide 2.1- 15 Using First Derivatives to Find Maximum and Minimum Values Example 3 (continued): 3rd x = 2 partitions the number line into 2 intervals: A (– ∞, 2) and B (2, ∞). So, analyze the signs of f (x) in both intervals. B A Interval x 2 Test Value Sign of f (x) Result x=0 x=3 – + f is decreasing f is increasing on on (–Pearson ∞, 2] [2, ∞) Copyright © 2008 Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 16 Using First Derivatives to Find Maximum and Minimum Values Example 3 (continued): Therefore, by the First-Derivative Test, f has a relative minimum at x = 2 given by f (2) (2 2)2 3 1 1 The relative minimum value is 1. It occurs at x = 2. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 17 Using First Derivatives to Find Maximum and Minimum Values Example 3 (concluded): We use the information obtained to sketch the graph below, plotting other function values as needed. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 18 More Examples Find the relative extrema. f ( x) x 1 3 x f ( x) x4 2 x 4 x 21 f ( x) x2 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 19 C(q) 25q 5000 p(q) 90 0.02 q C(q) above is the cost function. p(q) is the price function. Find a) The number of units that will produce a maximum profit. b) The maximum profit. c) The price that will produce a maximum profit. a) 1625 b) $47812.50 c) $57.50 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson AddisonWesley Slide 2.1- 20