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3.1 Extrema on an Interval Don’t get behind the Do your homework meticulously!!! Definition of Extrema 1. f(c) is the minimum of f on I if f(c) £ f(x) " x in I. 2. f(c) is the maximum of f on I if f(c) ! f(x) " x in I. If f is continuous on a closed interval [a,b], then f has both a minimum and a maximum on the interval. If f is defined at c, then c is called a critical number of f if f’(c) = 0 or if f’(c) is undefined at c. Relative Extrema occur only at critical numbers. If f has a relative min or relative max at x = c, then c is a critical number of f. Ex. Find the absolute extrema of f(x) = 3x4 – 4x3 on the interval[-1, 2]. f’(x) = 12x3 – 12x2 0 = 12x3 – 12x2 0 = 12x2(x – 1) x = 0, 1 are the critical numbers Evaluate f at the f(-1) = endpoints of [-1, 2] f(0) = and the critical #’s. f(1) = f(2) = 7 0 -1 16 minimum maximum Ex. Find the absolute extrema of f(x) = 2x – 3x2/3 on [-1, 3] 2 -1 3 f ' ( x) = 2 - 2 x = 2- 13 x ( ) 2 x1 3 - 2 2 x1 3 - 1 = = =0 13 13 x x C. N. ‘s are x = 1 because f’(1) = 0 and x = 0 because f’ is undefined Evaluate f at the endpoints of [-1, 3] and the critical #’s. f(-1) = -5 minimum f(0) = 0 maximum f(1) = -1 f(3) = -.24 Ex. Find the absolute extrema of f(x) = 2sin x – cos 2x on [0,2p ]. f’(x) = 2cos x + 2sin 2x sin 2x = 2 sin x cos x 0 = 2cos x + 2(2sin x cos x) Factor 0 = 2cos x(1 + 2sin x) p 3p @ and 2 2 7p 11p sin x = -1/2 @ and 6 6 Evaluate f at the endpoints of [0,2p ] and the critical #’s. 2cos x = 0 cos x = 0 f (0) = Êp ˆ fÁ ˜= Ë2¯ Ê 7p ˆ fÁ ˜= Ë 6 ¯ Ê 3p ˆ fÁ ˜= Ë 2 ¯ Ê 11p ˆ fÁ ˜= Ë 6 ¯ f (2p ) = -1 3 max -3 min 2 -1 -3 min 2 -1