Absolute Extrema. Joseph Lee. Metropolitan Community College

Transcription

1 Metropolitan Community College

2 Let f be a function defined over some interval I. An absolute minimum occurs at c if f (c) f (x) for all x in I. An absolute maximum occurs at c if f (c) f (x) for all x in I.

3 Extreme Value Theorem If f is a continuous function on a closed interval [a, b], then f possesses both an absolute minimum and an absolute maximum.

4 Relative Minima and Maxima Let c be a real number in the open interval (a, b). A relative minimum occurs at c if f (c) f (x) for all x in (a, b). A relative maximum occurs at c if f (x) f (x) for all x in (a, b).

5 Critical Numbers Let f (c) be defined. Then c is called a critical number of f if f (c) = 0 or f (c) is undefined.

6 Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6

7 Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6 f (x) = 2x 1

8 Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6 f (x) = 2x 1 0 = 2x 1

9 Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6 f (x) = 2x 1 0 = 2x = x

10 Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6 f (x) = 2x 1 0 = 2x 1 f ( 3) = 6 f 1 2 = x ( ) 1 = f (3) = 0

11 Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6 f (x) = 2x 1 0 = 2x = x ( ) 1 f ( 3) = 6 f = 25 f (3) = ( ) 1 Absolute Minimum: 2, 25 Absolute Maximum: (3, 6) 4

12 Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x

13 Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x

14 Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x 0 = cos x

15 Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x 0 = cos x x = π 2, 3π 2

16 Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x 0 = cos x g(0) = 0 x = π 2, 3π 2 ( π ) ( ) 3π g = 1 g = 1 g(2π) = 0 2 2

17 Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x 0 = cos x g(0) = 0 Absolute Minimum: ( π ) g = 1 g 2 ( ) 3π 2, 1 x = π 2, 3π 2 ( ) 3π = 1 g(2π) = 0 2 ( π ) 2, 1 Absolute Maximum:

18 Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x x

19 Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x x h(x) = x + 1 x

20 Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x x h(x) = x + 1 x h (x) = 1 1 x 2

21 Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x x h(x) = x + 1 x h (x) = 1 1 x 2 0 = 1 1 x 2

22 Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x x h(x) = x + 1 x h (x) = 1 1 x 2 0 = 1 1 x 2 x = ±1

23 Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x x h(x) = x + 1 x h (x) = 1 1 x 2 0 = 1 1 x 2 h ( ) 1 = x = ±1 h(1) = 2 h(5) = 26 5

24 Example 3. (continued) Find the absolute extrema on the closed interval [ 1 2, 5]. h ( ) 1 = h(x) = x x h(1) = 2 h(5) = 26 5

25 Example 3. (continued) Find the absolute extrema on the closed interval [ 1 2, 5]. h ( ) 1 = Absolute Minimum: (1, 2) h(x) = x x h(1) = 2 h(5) = 26 5 ( Absolute Maximum: 5, 26 ) 5

26 Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3

27 Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3 k(x) = { x 3 x 3 (x 3) x < 3

28 Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3 k(x) = k (x) = { x 3 x 3 (x 3) x < 3 { 1 x > 3 1 x < 3

29 Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3 k(x) = k (x) = { x 3 x 3 (x 3) x < 3 { 1 x > 3 1 x < 3 k (x) 0

30 Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3 k(x) = k (x) = { x 3 x 3 (x 3) x < 3 { 1 x > 3 1 x < 3 k (x) 0 k(0) = 3 k(3) = 0 k(5) = 2

31 Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3 k(x) = k (x) = { x 3 x 3 (x 3) x < 3 { 1 x > 3 1 x < 3 k (x) 0 k(0) = 3 k(3) = 0 k(5) = 2 Absolute Minimum: (3, 0) Absolute Maximum: (0, 3)