Metropolitan Community College
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.
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.
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).
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.
Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6
Example 1. Find the absolute extrema on the closed interval [ 3, 3]. f (x) = x 2 x 6 f (x) = 2x 1
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
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 1 2 = x
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 = 25 2 4 f (3) = 0
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 1 2 = x ( ) 1 f ( 3) = 6 f = 25 f (3) = 0 2 4 ( ) 1 Absolute Minimum: 2, 25 Absolute Maximum: (3, 6) 4
Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x
Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x
Example 2. Find the absolute extrema on the closed interval [0, 2π]. g(x) = sin x g (x) = cos x 0 = cos x
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
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
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:
Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x 2 + 1 x
Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x 2 + 1 x h(x) = x + 1 x
Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x 2 + 1 x h(x) = x + 1 x h (x) = 1 1 x 2
Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x 2 + 1 x h(x) = x + 1 x h (x) = 1 1 x 2 0 = 1 1 x 2
Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x 2 + 1 x h(x) = x + 1 x h (x) = 1 1 x 2 0 = 1 1 x 2 x = ±1
Example 3. Find the absolute extrema on the closed interval [ 1 2, 5]. h(x) = x 2 + 1 x h(x) = x + 1 x h (x) = 1 1 x 2 0 = 1 1 x 2 h ( ) 1 = 5 2 2 x = ±1 h(1) = 2 h(5) = 26 5
Example 3. (continued) Find the absolute extrema on the closed interval [ 1 2, 5]. h ( ) 1 = 5 2 2 h(x) = x 2 + 1 x h(1) = 2 h(5) = 26 5
Example 3. (continued) Find the absolute extrema on the closed interval [ 1 2, 5]. h ( ) 1 = 5 2 2 Absolute Minimum: (1, 2) h(x) = x 2 + 1 x h(1) = 2 h(5) = 26 5 ( Absolute Maximum: 5, 26 ) 5
Example 4. Find the absolute extrema on the closed interval [0, 5]. k(x) = x 3
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
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
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
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
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)