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Section 3.1 Extrema on an Interval Definition: Extrema Let f be defined on an interval I containing c. f (c) is the minimum of f on I if f ( x) ≥ f (c) for all x in I f (c) is the maximum of f on I if f ( x) ≤ f (c) for all x in I The minimum and maximum of a function on an interval are the extreme values, or extrema, of the function on the interval. The minimum and maximum of a function on an interval are also called the absolute minimum and absolute maximum on the interval. Theorem: The Extreme Value Theorem If f is continuous on a closed interval [a, b], then f has both a minimum and a maximum on the interval. Definition: Relative Extrema If there is an open interval containing c on which f (c) is the minimum, then f (c) is called a relative minimum of f. If there is an open interval containing c on which f (c) is the maximum, then f (c) is called a relative maximum of f. Calculus I by Chinyoung Bergbauer at NHMCCD: 3.1 Extrema on an Interval 1 Example 1: Find the value of the derivative at each of the relative extrema of the given function. a) f ( x) = ( x − 1)( x + 2)( x − 3) The function has 2 relative extrema, one relative maximum and one relative minimum. The derivatives at the relative extrema are 0. b) f ( x) =| x − 1 | The function has only one relative minimum at 1 and its derivative at 1 is undefined. Calculus I by Chinyoung Bergbauer at NHMCCD: 3.1 Extrema on an Interval 2 c) f ( x) = cos x The function has infinitely many relative extrema. The function has relative maxima at x = 2nπ and relative minima at x = (2n + 1)π . Definition: Critical Number Let f be defined at c. If f ′(c) = 0 or if f is not differentiable at c, then c is a critical number of f. Theorem: Extrema occur only at Critical Numbers If f has a relative maximum or minimum at c, then c is a critical number of f. How to find (Absolute) Extrema of a continuous function on a closed interval [a,b] 1. Find the critical numbers c on the interval. 2. Evaluate all f (c) ’s, f (a) , and f (b) . 3. The least of these values is the minimum. The greatest is the maximum. Calculus I by Chinyoung Bergbauer at NHMCCD: 3.1 Extrema on an Interval 3 Example 2: Find the extrema on a closed interval. f ( x) = 16 x 4 − 8 x 2 on [-1, 1] Step 1: Find the critical numbers: f ( x) = 16 x 4 − 8 x 2 f ′( x) = 64 x 3 − 16 x = 16 x(4 x 2 − 1) = 16 x(2 x − 1)(2 x + 1) = 0 when x = 0,1 / 2,−1 / 2 f (−1) = 8 f ( 0) = 0 f (−1 / 2) = −1 f (1 / 2) = −1 f (1) = 8 The function has the absolute minimum -1 at x = ½ and -1/2. The function has the absolute maximum 8 at x = 1 and x = -1. Practice 2: Find the extrema on a closed interval. f ( x) = t on [3, 5] t−2 Example 3: Find the extrema of f ( x) = 4 x − 63 x 2 on [-1, 2] f ( x ) = 4 x − 63 x 2 = 4 x − 6 x 2 / 3 f ′( x) = 4 − 4 x −1/ 3 = 4 − 3 4 43 x − 4 = 3 x x f ′( x) = 0 when 43 x − 4 = 0 ⇒ 4(3 x − 1) = 0 ⇒ x = 1 : critical number f ′( x) is undefined when 3 x = 0 ⇒ x = 0 :: critical number f ( 0) = 4( 0) − 6 3 0 2 = 0 f (1) = 4 − 63 12 = −2 f (−1) = −4 − 63 (−1) 2 = −10 f (2) = 8 − 63 2 2 = 8 − 63 4 ≈ −1.5244 The function has the maximum 0 at x = 0 and the minimum -10 at x = -1. Example 4: Find the extrema of f ( x) = sin 2 x + 2 cos x on [0,2π ] Calculus I by Chinyoung Bergbauer at NHMCCD: 3.1 Extrema on an Interval 4 f ( x) = sin 2 x + 2 cos x f ′( x) = 2 cos 2 x − 2 sin x = 2(1 − 2 sin 2 x) − 2 sin x = −4 sin 2 x − 2 sin x + 2 Set f ′(x) = 0 to find the critical numbers. Note that f ′(x) is always defined. − 4 sin 2 x − 2 sin x + 2 = 0 2 sin 2 x + sin x − 1 = 0 (2 sin x − 1)(sin x + 1) = 0 1 or sin x = −1 2 π 5π 3π x = , ,x = : critical numbers 2 6 6 π π 3 3 π f = sin + 2 cos = : max 3 6 2 6 sin x = 3 3 3 5π 3 5π 5π f = sin + 2 cos =− −2 =− : min 2 2 6 2 3 6 3π 3π f = sin 3π + 2 cos =0 2 2 f (0) = sin 0 + 2 cos 0 = 2; f (2π ) = sin 4π + 2 cos 2π = 2 Practice 3: Find the extrema of f ( x) = 33 x 2 − 2 x on [−1,1] Practice 4: Find the extrema of f ( x) = x 2 − 2 − cos x on [−1,3] Calculus I by Chinyoung Bergbauer at NHMCCD: 3.1 Extrema on an Interval 5 Calculus I by Chinyoung Bergbauer at NHMCCD: 3.1 Extrema on an Interval 6