Martens_WA4 - 4.1 24 f ( x) = x 2 + 2 x − 4,[−1,1] f '(...

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Unformatted text preview: 4.1 24 f ( x) = x 2 + 2 x − 4,[−1,1] f '( x) = 2 x + 2 2x + 2 = 0 2(−1) + 2 = 0 -1 is an end point and hence not a critical number. f ( −1) = −12 + 2(−1) − 4 f ( −1) = 1 − 2 − 4 f ( −1) = −5 Endpoints [-1,1] f (1) = (1) 2 + 2(1) − 4 f (1) = 1 + 2 − 4 f (1) = −1 Left endpoint: f (−1) = −5 Right endpoint: f (1) = −1 Min Max 28 g ( x) = 3 x ,[−1,1] g ( x) = ( x) 3 1 g '( x) = x −2/3 (1) 3 1 −2/3 1 g '( x) = x = 3 3 3 x2 g '( x) = g '(0) = 1 3 x2 1 3 3 02 3 1 = undefined Critical point = 0 3 Left endpoint g (−1) = −1 = −1 3 Right endpoint: g (1) = 1 = 1 3 Critical point: g (0) = 0 = 0 Min Max 4.4 12 f ( x) = 2 x3 − 3 x 2 − 12 x + 5 f '( x) = 6 x 2 − 6 x − 12 f ''( x) = 12 x − 6 12 x − 6 = 0 12(.5) − 6 = 0 −∞ < 0 < .5 .5 < 1 < ∞ f ''(0) = 12(0) − 6 = −6 f ''(1) = 12(1) − 6 = 6 Concave down. Concave up. Point of inflection at [.5, −1.5] 14 f ( x) = 2 x 4 − 8 x + 3 f '( x) = 8 x 3 − 8 f ''( x ) = 24 x 2 f ''( x ) = 24 x 2 = 0 f ''( x ) = 24(0) 2 = 0 Critical point = 0 Test values: −∞ < −1 < 0 0 <1< ∞ f ''(−1) = 24(−1) 2 = 24 f ''(1) = 24(1) 2 = 24 Concave up. Concave up. No points of inflection because it is U shaped concave up, concave up. 4.5 22 3x3 + 2 lim 3 x →∞ 9 x − 2 x 2 + 7 2 x 3 (3 + 3 ) x lim x →∞ 3 27 x (9 − + 3 ) xx 2 3+ 3 31 x lim == x →∞ 27 9− + 3 9 3 xx 30 lim −3 x + 1 x2 + x x →∞ 1 x ( −3 + ) x lim x →∞ x x 2 (1 + 2 ) x 1 x( −3 + ) x − 1 −3 = 3 lim x →∞ 1 1 | x | 1+ x 64 y= y' = x −3 x−2 ( x − 2)(1) − ( x − 3)(1) ( x − 2) 2 ( x − 2) − ( x − 3) y' = x2 − 4 x + 4 1 1 y' = 2 = x − 4 x + 4 ( x − 2)2 Extrema: 1 ( x − 2) 2 1 f '(2) = = undefined (2 − 2) 2 f '( x) = However f ( 2) is undefined as well so there is no extrema. Intercepts: x −3 x−2 x −3 0= x−2 y= 0= 3−3 0 = 3− 2 1 x-intercepts [3,0] x −3 x−2 −3 1 y= =1 −2 2 y= 1 [0,1 ] 2 y-intercept Symmetry: y= x−3 x−2 x−3 f ( x) = x−2 −x − 3 −x − 2 x−3 f ( − x) = − = − f ( x) x−2 f ( − x) = f ( x) is symmetric with respect to a line on the graph. Asymptotes: 2 − 3 −1 = 2−2 0 Asymptote at: x=2 4.6 16 f ( x) = x3 x2 − 4 ( x 2 − 4)(3 x 2 ) − ( x 3 )(2 x ) x 4 − 12 x 2 x 4 − 12 x 2 =2 =4 ( x 2 − 4) 2 ( x − 4) 2 x − 8 x 2 + 16 f '( x) = ( x 4 − 8 x 2 + 16)(4 x 3 − 24 x) − ( x 4 − 12 x 2 )(4 x 3 − 16 x) f ''( x) = ( x 4 − 8 x 2 + 16) 2 (4 x 7 − 24 x 5 − 32 x 5 + 192 x 3 + 64 x 3 − 384 x ) − (4 x 7 − 16 x 5 − 48 x 5 + 192 x 3 ) f ''( x) = ( x 4 − 8 x 2 + 16) 2 (8 x 5 + 64 x 3 − 384 x) (8 x 5 + 64 x 3 − 384 x ) f ''( x) = =8 ( x 4 − 8 x 2 + 16) 2 x − 16 x 6 + 96 x 4 + 256 Intercepts: f ( x) = y= x3 x2 − 4 x3 x2 − 4 03 0 y= 2 = =0 0 − 4 −4 y-intercept [0, 0] (note: x-intercept will be the same.) f ( x) = 0= x3 x2 − 4 x3 x2 − 4 03 0 0= 2 = =0 0 − 4 −4 x-intercept [0, 0] Asymptote: x3 f ( x) = 2 x −4 3 2 8 8 = = 2 2 −4 4−4 0 23 8 8 = = 2 −2 − 4 4 − 4 0 Vertical Asymptote at + − 2 No horizontal asymptote because the degree is larger in the numerator. Critical Values: x 4 − 12 x 2 x 4 − 12 x 2 f '( x ) = 2 = ( x − 4) 2 x 4 − 8 x 2 + 16 f '( x) = f '( x) = x 4 − 12 x 2 =0 ( x 2 − 4) 2 24 − 12(2) 2 = undefined (22 − 4) 2 x = 2 and x = −2 are NOT critical values because the asymptote is at 2 and -2. x 4 − 12 x 2 f '( x) = 2 =0 ( x − 4) 2 f '( x) = 04 − 12(0) 2 =0 (02 − 4) 2 x = 0 is a critical value. f '( x ) = x 4 − 12 x 2 ( x 2 − 4) 2 x 4 − 12 x 2 = 0 ( x 2 )( x 2 − 12) = 0 ( x2 ) = 0 ( x 2 − 12) = ± 12 x = 12 is a critical value. x = − 12 is a critical value. Relative Extrema: 12, − 12, 0 f ( − 12) = (− 12)3 (− 12) 2 − 4 12 − 12 (4)(3)(2) − 3 (8)(3) − 3 = = = 3 − 3 = − 27 8 8 8 f ( − 12) = Reletive Max f ( 12) = ( 12)3 ( 12) 2 − 4 12 12 4 12 (4)(3)(2) 3 (8)(3) 3 =1 = = 3 3 = 27 8 8 8 8 f ( 12) = Reletive Min f (0) = (0)3 =0 (0) 2 − 4 Points of inflection: 64 384 − ) x2 x4 f ''( x) = 96 256 x 5 ( x3 − 16 x + + 5 ) x x x5 (8 + We have possible points of inflection at 0 Test values for 0: −∞ < −1 < 0 0 <1< ∞ With test value -1. (8(−1)5 + 64( −1)3 − 384(−1)) ( −1)8 − 16(−1)6 + 96(−1) 4 + 256 (−8 − 64 + 384) 312 f ''(−1) = = 1 − 16 + 96 + 256 337 f ''(−1) = With test value 1. (8(1)5 + 64(1)3 − 384(1)) (1)8 − 16(1)6 + 96(1) 4 + 256 (8 + 64 − 384) −312 f ''(1) = = 1 − 16 + 96 + 256 367 f ''(1) = f ''(−1) > 0 is concave upward. f ''(1) < 0 is concave downward. is a point of inflection. Concavity: Find values for: f ''(0) = 12 12 f ''( 12) = 3 96 256 12 − 16 12 + + 5 12 12 64 384 8+ − 12 144 f ''( 12) = 96 256 12 12 − 16 12 + + 12 144 12 1 2 2 8+5 −2 10 3 3 3 f ''( 12) = = (96)144 12 + 256( 12) 12 12(96(9) + 256(16)) −4 12 + −4 12 + 144(12) 12*12 12 12 2 2 2 10 10 10 3 3 3 = = = 4960 − 576 4384 4960 4 12 − 12 12 12 12 1 12 12 10.66666 ≈ = .101141647 105.462649 8+ 64 12 2 − 384 4 2 < 12 < ∞ concave up −∞ < 12 < −2 concave down 34 f ( x) = x 4 − 8 x3 + 18 x 2 − 16 x + 5 f ( x) = x ( x 3 − 8 x 2 + 18 x − 16) + 5 f '( x ) = 4 x 3 − 24 x 2 + 36 x − 16 f ''( x) = 12 x 2 − 48 x + 36 Intercepts : f ( x) = x ( x 3 − 8 x 2 + 18 x − 16) + 5 f (0) = 0(03 − 8(0) 2 + 18(0) − 16) + 5 f (0) = 5 y -intercept [0,5] f ( x) = x 4 − 8 x3 + 18 x 2 − 16 x + 5 0 = 1(13 − 8(1) 2 + 18(1) − 16) + 5 0 = 1 − 8 + 18 − 16 + 5 = −7 + 2 + 5 = −7 + 7 = 0 x -intercept [1, 0] f ( x) = x 4 − 8 x3 + 18 x 2 − 16 x + 5 0 = x 2 ( x 2 − 8 x + 18) − 16 x + 5 0 = 25(25 − 40 + 18) − 16(5) + 5 0 = 25(−15 + 18) − 75 0 = 25(3) − 75 = 75 − 75 x-intercept [5, 0] Relative E xtrema: f '( x) = 4 x 3 − 24 x 2 + 36 x − 16 0 = 4 x 3 − 24 x 2 + 36 x − 16 0 = 4(1)3 − 24(1) 2 + 36(1) − 16 0 = 4 − 24 + 36 − 16 Critical point at x = 1 f ( x) = x( x 3 − 8 x 2 + 18 x − 16) + 5 f (1) = 1(13 − 8(1) 2 + 18(1) − 16) + 5 f (1) = 1(1 − 8 + 18 − 16) + 5 f (1) = −7 + 2 + 5 f (1) = 0 Critical point: [1, 0] -- f '( x ) = 4 x 3 − 24 x 2 + 36 x − 16 0 = 4 x 3 − 24 x 2 + 36 x − 16 0 = 4(4)3 − 24(4) 2 + 36(4) − 16 0 = 4(64) − 24(16) + 36(4) − 16 0 = 256 − 384 + 144 − 16 0 = −128 + 128 Critical point at [4, 0] . f ( x) = x( x3 − 8 x 2 + 18 x − 16) + 5 f (4) = 4(43 − 8(4) 2 + 18(4) − 16) + 5 f (4) = 4(64 − 128 + 72 − 16) + 5 f (4) = −32 + 5 f (4) = −27 Critical point at : [4, −27] Reletive min. Points of inflection f ''( x) = 12 x 2 − 48 x + 36 0 = 12 x 2 − 48 x + 36 0 = 12( x 2 − 4 x + 3) 0 = 12((1 − x)(3 − x)) 0 = 12((1 − x )(3 − 3)) 0 = 12((1 − x )(0)) 0 = 12((1 − 1)(3 − 1)) 0 = 12((0)(3 − 1)) Possible points of inflection at 3 and 1. Test values: −∞ < 0 < 1 1< 2 < 3 3< 4<∞ f ''( x) = 12((1 − x)(3 − x)) f ''(0) = 12((1 − 0)(3 − 0)) = 36 f ''(0) is concave up f f f f f ''( x) = 12((1 − x)(3 − x)) ''(2) = 12((1 − 2)(3 − 2)) ''(2) = 12((−1)(1)) ''(2) = 12(−1) ''(2) = −12 f ''(2) is concave down. f ''( x) = 12((1 − x)(3 − x)) f ''(4) = 12((1 − 4)(3 − 4)) f ''(4) = 12((−3)(−1)) f ''(4) = 12(3) = 36 f ''(4) is concave up. Asymptotes There are none. 4.7 2(a-c) Height 1 2 3 4 5 6 Length and width Volume 24 − 2 ( 1) 24 − 2 ( 2 ) 24 − 2 ( 3) 24 − 2 ( 4 ) 24 − 2 ( 5 ) 24 − 2 ( 6 ) 1 24 − 2 ( 1) = 484 2 2 24 − 2 ( 2 ) = 800 2 3 24 − 2 ( 3) = 972 2 4 24 − 2 ( 4 ) = 1024 2 5 24 − 2 ( 5 ) = 980 2 6 24 − 2 ( 6 ) = 864 2 v = x 24 − 2 ( x ) v = x 24 − 2 ( x ) 2 2 v = 4( x 3 − 24 x 2 + 144 x) v ' = x 2 24 − 2 ( x ) ( −2) + 24 − 2 ( x ) (1) 2 v ' = 12( x 2 − 16 x + 48) Critical Values: 4,12 v '' = 12(2 x − 16) v '' = 24( x − 8) v ''(4) = 24(4 − 8) < 0 concave down v ''(4) gives reletive max. v ''(12) = 24(4 − 8) > 0 concave up 8 x 2 + y = 27 −27 + x 2 = − y y = 27 − x 2 f ( x) = xy f ( x) = x (27 − x 2 ) f ( x) = x 27 − x 3 f '( x ) = 27 − 3 x 2 0 = 27 − 3 x 2 3 x 2 = 27 x2 = 9 32 = 9 y = 27 − 32 y = 27 − 9 y = 18 Numbers 3 and 18. 20 Perimeter= P=200 P=4x+3y 200=4x+3y Area= A = 2 xy -- 200 = 4 x + 3 y −3 y + 200 = 4 x −3 y + 200 =x 4 3y 50 − =x 4 A = 2 xy 3y A = 2 y 50 − 4 2 y 3y A = y100 − * 14 6 y2 A = y100 − 4 3 A = y100 − y 2 2 A ' = 100 − 3 y 0 = 100 − 3 y 100 y= 3 A '' = −3 Always concave down. Therefore y = 100 gives max value. 3 100 200 = 4 x + 3 3 300 200 = 4 x + 3 200 = 4 x + 100 100 = 4 x 25 = x The corrals should be 25 foot wide each and have a length of 100/3 in order to maximize the area. 30. A = ( x + 3)( y + 3) A = xy + 3 x + 3 y + 9 36 = xy 36 y= x 36 A = 36 + 3 x + 3 + 9 x 108 A = 45 + 3 x + x 108 x2 A ' = 3 x 2 − 108 A' = 3− A ' = x 2 − 36 x 2 = 36 x = ±6 Critical numbers at ±6 . -6 is out of the domain because it’s negative and we’re looking for numbers >0 A = 45 + 3(6) + 108 6 A = 45 + 18 + 18 A = 81 81 = 13.5 6 y = 13.5 x=6 ...
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This note was uploaded on 12/30/2009 for the course MAT 231 taught by Professor Thurber during the Spring '09 term at Thomas Edison State.

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