Answers to review exercises(2)

Answers to review exercises(2) - 导导导导...

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Unformatted text preview: 导导导导 (导导导导导导导 p147-150 1 ⇔ (( 3.( .(((((( ∴ 2 x = −3 x , x0 = 0(( 2 k1 = k 2 , 2 −. 3 4(1) f ' ( x) = (14)' ≡ 0,∴ f ' (− 2) = f ' (2) = 0. 1 n 1 n m −1 4( 2) y ' = 2 x, ,− ,x. 2x 2x x m 1 4(3)(( (log a x )' = , x ln a 1 1 −1 1 y' = , , ,. x ln 10 x ln 2 x ln 2 x a x )' = a x ln a, 1x x x y ' =10 ln 10,2 ln 2,− ) ln 2, e x . ( 2 2 4( 4)((( 4 3 4 5(1) y = 2 x − 3 x + , y ' = 4 x − − 2. x 2x x 2 5( 2) y = 2 − ln x 1 x ; y ' = 2 ln 2 − 2 x. x x 2 −1 5(3) y = ; x+2 ( x 2 − 1)' ( x + 2) − ( x 2 − 1)( x + 2)' 2 x( x − 2) − ( x 2 − 1) ⋅1 y' = = 2 ( x + 2) ( x + 2) 2 x2 + 4x +1 = . 2 ( x + 2) sin x (sin x)' x − sin x( x)' x cos x − sin x 5(4) y = ; y' = = . 2 2 x x x 3 5(5) y = ( x −1)(2 x −1) 2 ; y ' = ( x −1)' ( 2 x −1) 2 + ( x −1)[(2 x −1) 2 ]' = ( 2 x −1) 2 + 4( x −1)(2 x −1) = 12 x 2 −16 x + 5. 5(6) y = x ln x. ln x x 1 ln x y ' = ( x )' ln x + x (ln x)' = + = (1 + ). x 2 2x x x 6(1) y = e . x e ( . y ' = e ⋅ ( x )' = . 2x x 6( 2) y = sin e x y ' = cos e ⋅ (e )' = e cos e . x x x x 4 1 − ln x 2 6(3) y = ( ) 1 + ln x 1 1 − (1 + ln x) − (1 − ln x) 1 − ln x x y ' = 2( )⋅ x 2 1 + ln x (1 + ln x) 4(ln x − ) 1 = . 3 x (1 +ln x ) x −1 3 6( 4) y = ( ). x +1 x −1 2 2 6( x − 1) y ' = 3( )⋅ = . 2 3 x + 1 ( x + 1) ( x + 1) 2 5 y = x ln x 2 − 1 1 y ' = ln x − 1 + x ⋅ [ln( x 2 − 1)]' 2 2 x x 2x 2 2 = ln x − 1 + 2 . = ln x − 1 + ⋅ 2 x −1 2 x −1 2 6(6) y = 10 y ' = 10 sin 1 x sin 1 x . 1 1 ln 10 ⋅ cos ⋅ (− 2 ) x x 6 6(7) y = tan ln x. 1 1 y' = ⋅ (ln x)' = . 2 2 cos ln x x cos ln x 6(8) y = x 1 − sin x 3 y ' = ( x)' 3 1 − sin x + x ⋅ (3 u )' (1 − sin x)' 2 3 1 = 1 − sin x + x ⋅ (1 − sin x) (− cos x) 3 x cos x 3 = 1 − sin x [1 − 3(1 − sin x) 3 − 7 6(9) y = arcctgx 2 ; ⇒ y = arcctgu , u = x 2 . − 2x y ' = (arcctgu )' ( x )' = . 4 1+ x 2 6(10) y = e y' = e =e arcsin x ⋅ (arcsin x )' 1 1 1 x arcsin x ⋅ ⋅ = e . 1 − x 2 x 2 x(1 − x) arcsin x arcsin 8 6(11) y = ln( x + 1 + x 2 ) + x cos 2 x. 1 2 ( .[ln( x + 1 + x )]' = ((( p81(10.8)) 1+ x2 ( x cos x)' = cos x + x ⋅ 2 cos x ⋅ (− sin x). 1 ∴(( = + cos 2 x − x sin 2 x. 1+ x2 6(12) y = ( x − a)( x − b)( x − c)( x − d ) 1 ln y = [ln( x − a ) + ln( x − b) + ln( x − c) + ln( x − d )] 2 y1 1 1 1 y' = ( + + + ). 2 x −a x −b x −c x −d 9 2 2 7(1) ln x +y 2 2 y = arctan , ( y ' x ( .(((( x( ( 1 1 ( xy '− y ) (2 x + 2 yy ' ) = ⋅ . 2 2 2 y2 2( x + y ) x 1+ ( ) x x+ y ( x + yy ' = xy '− y,∴ y ' = . x− y 7(2)e x+ y ( .(((( dy − xy = 1, ( | x =0. dx x( ( : e x + y (1 + y ' ) − ( y + xy ' ) = 0, y − e x+y y' = x+y ,∴y (0) = −1. e −x 10 8.(((((( ( 1 y = x 4 − 5 x 3 + 6 x − 4. ( y ' = 4 x 3 − 15 x 2 + 6, y ' ' = 12 x 2 − 30 x (2) y = ln(1 + x). 1 1 y' = , y' ' = − . 2 1+ x (1 + x) −x (3) y = xe . −x −x −x y ' = e + x(e )' = e (1 − x), −x −x −x y ' ' = (e )' (1 − x) + e (1 − x)' = e ( x − 2). 11 x ( 4) y = . 1+ x 1 ⋅ (1 + x) − x ⋅1 −2 −2 y' = = (1 + x) , y ' ' = . 2 3 (1 + x) (1 + x) (1) y = e 9.(((( y' = e cos 1 x cos 1 x . 1 1 1 ⋅ (cos )' = 2 sin e x x x 1 1 ∴ = 2 sin e dy x x cos 1 x cos 1 x , dx. 12 1+ x (2) y = ln . 1− x 1 (−1) 2 y ' = [ln(1 + x) − ln(1 − x)]' = − = 1+ x 1− x 1− x2 2 ∴= dy dx. 2 1 −x −x (3) y = e cos 2 x. −x −x −x y ' = (− e ) cos 2 x + e (− 2 sin 2 x) = − e (cos 2 x + 2 sin 2 x), ∴ dy = −e − x (cos 2 x + 2 sin 2 x)dx 13 1+ x2 9( 4) y = . 2 1− x 2 2 2 x(1 − x ) − (1 + x )(−2 x) 4x y' = = , 22 22 (1 − x ) (1 − x ) 4x ∴ dy = dx. 22 (1 − x ) 2 2 x y (5) 2 + 2 = 1.(((( x((( ab 2 2 bx 2 x 2 yy ' bx dy = − 2 dx + 2 = 0, y ' = − 2 . 2 ay a b ay 14 9(6) y = 1 − xe , ((( 2 y x((( −e 2 yy ' = −(e + xe y ' ),∴ y ' = . y 2 y + xe y −e dy = dx. y 2 y + xe y y y − 3 −1 10.( AB((( k= = −2, 3 −1 (( // AB ⇔ f ' ( x) = k 2 − 2 x = −2,∴ x = 2, y = 0. (( 2(0((((((((( AB. 15 (1 − x )α −1 11(1) lim . x →0 x α −1 −α (1 − x ) (( = lim = −α. x →0 1 sin 2 x (2) lim . x→0 x 2 cos 2 x (( = lim = 2 cos 0 = 2. x→0 1 ln sin x cos x (3) lim = lim 2 π π x → (π − 2 x ) x → − 4 sin x (π − 2 x ) 2 2 − sin x 1 = lim =− π 1 8 x → − 4[cos x (π − 2 x ) − 2 sin x ] 6 2 cot x − csc x −x (4) lim+ = lim+ = lim+ 2 x →0 ln x x →0 x →0 sin x 1/ x 2 2 −1 = lim =− ∞ + x →0 sin 2 x ln x 2 ln x 2 (5) lim = lim = lim = 0 x→ + ∞ x x→ + ∞ x x→ + ∞ x ln(1 − x ) −1 (6) lim = lim = −1 x →0 x →0 1 − x x 17 1 3 x + x −2 (7) lim( −3 ) = lim 3 x → x −1 1 x→ 1 x −1 x −1 2 x +1 = lim =1 2 x→ 1 3x 1 1 ln x − ( x − 1) (8) lim( − ) = lim x →1 x − 1 x →1 ( x − 1) ln x ln x 1 −1 1− x x = lim = lim x →1 x − 1 x→1 x ln x + x − 1 ln x + x 2 −1 1 = lim =− . x → ln x + 2 1 2 18 12(1) f ( x) = x − 3 x −1 3 f ' ( x) = 3 x − 3, x0 = ±1 2 x f’(x) (−∞ ,− 1) + -1 (-1,1) (1,+ ∞) 1 - + f(x) 12(2) f ( x) = x 3 + 3 x 2 + 3 x − 4. f ' ( x) = 3 x 2 + 6 x + 3 = 3( x + 1) 2 ≥ 0, ∴ f ( x)( (− ,+ ∞ ((((( ∞ ) 19 12(3) f ( x) = x − x, 1 1 f ' ( x) = − 1, x0 = 0, . 4 2x x f’(x) (0,1/4) + 1/4 1 ( ,+ ∞ ) 4 _ f(x) 20 x −1 12(4) f ( x) = , x +1 2 f ' ( x) = ≥ 0,∴ f ( x)(( 2 ( x + 1) − ∞ ,+ ∞)((((( 12(5) f ( x ) = x −arctan x, 1 x2 f ' ( x ) =1 − = ≥ 0, 2 2 1+x 1+x ∴ f ( x )( ( − ,+ ∞((((( ∞ ) . 21 . 2 x 12(6) f ( x) = , 1+ x x ( x + 2) f ' ( x) = , x0 = 0,2 2 (1 + x) x f’(x) (−∞ ,− 2) + -2 (-2,0) 0 (0,+ ∞) _ + f(x) 22 13(1) f ( x) = 2 x + 3 x + x + 2, 3 2 1 f ' ' ( x) = 6(2 x + 1) = 0, x0 = − , 2 1 (−∞ ,− ) 2 f’’(x) _ x f(x) 1 (− ,+ ∞) 2 + 1 − 2 导导导 -1/2 导 2 导 23 13( 2) f ( x ) = x −4 x +16, f ' ' ( x ) =12 x ( x −2) = 0, x0 = 0,2 4 x (−∞ ,0) + f’’(x) f(x) 0 3 (0,2) 2 (2,+ ∞) _ 导导导 0 导 16 导 + 导导 导2导0导 24 14. y = x 3 + ax 2 − 9 x + 4( x = 1(((( ( 1 y ' ' = 6 x + 2a, y ' ' (1) = 0,∴ 6 + 2a = 0, a = −3. ( (2) y = x 3 − 3x 2 − 9 x + 4, ( x = 1((( (((( y = −7, 1( − 7( (3) y ' = 3 x − 6 x − 9 = 3( x + 1)( x − 3), x0 = −1,3. 2 y ' ' = 6( x − 1), x1 = 1. 25 x (−∞ ,−1) f’(x) + _ _ f’’(x) _ _ + + f(x) -1 (-1,1) 1 (1,3) 3 (3,+ ∞ ) + 26 15(1) f ( x) = x 4 − 2 x 2 + 3, f ' ( x) = 4 x − 4 x = 4 x( x − 1) = 0, x0 = 0,±1 3 2 f ' ' ( x) = 12 x − 4, f ' ' (0) < 0, f ' ' ( ±1) > 0. 2 ((( f (0) = 3, ((( f ( ±1) = 2. 15(2) f ( x) = x 3 + 3x 2 − 24 x + 12, f ' ( x) = 3x + 6 x − 24 = 3( x + 2 x − 8) = 0, x0 = −4,2. 2 2 f ' ' ( x) = 6( x + 1), f ' ' (−4) < 0, f ' ' (2) > 0, ((( f( − 4( = 92(((( f (2) = −16 27 x4 (3) g ( x ) = + , 3x 1 4 g ' ( x ) = − 2 = 0 ⇒ x0 = ±2 3. 3x 8 g ' ' ( x) = 3 , g ' ' (2 3 ) > 0, g ' ' (−2 3 ) < 0. x 43 43 ((( g (2 3 ) = ; ((( g (−2 3 ) = − . 3 3 16 16 2 (4) g ( x) = x + , g ' ( x) = 2 x − 2 = 0, x0 = 2. x x 32 g ' ' ( x) = 2 + 3 , g ' ' ( 2) > 0.((( x g ( 2) = 12. 28 x2 + x +1 x2 −1 1 (5) f ( x) = , f ' ( x) = 2 = 1 − 2 = 0, x0 = ±1 x x x 2 f ' ' ( x ) = 3 , f ' ' (−1) < 0, f ' ' (1) > 0. x ((( f (−1) = −1, ((( f (1) = 3. 6x 6(1 − x 2 ) ( 6) f ( x ) = 2 , f ' ( x) = 2 = 0, x0 = ±1, 2 x +1 ( x + 1) −12 x(3 − x 2 ) f ' ' ( x) = , f ' ' ( −1) > 0, f ' ' (1) < 0. 2 3 ( x +1) ((( f (−1) = −3, ((( f (1) = 3. 29 16(1) f ( x) = x 5 − 5 x 4 + 5 x 3 + 1.x ∈ [−1,2]. f ' ( x) = 5 x ( x − 4 x + 3) = 0, x0 = 0,1,3(((( 2 2 f (0) = 1, f (1) = 2, f (− 1) = − 10, f (2) = − 7. ∴ M = 2, m = − 10. x2 1 x ( x + 2) 2) g ( x ) = , x ∈ [− ,1].g ' ( x) = = 0, x0 = 0,−(((( 2 2 1+ x 2 (1 + x) 1 1 1 g (0) = 0, g (− ) = , g (1) = , 2 2 2 1 ∴M = , m = 0. 2 30 (3) h( x ) = x + x , x ∈ 0,4]. [ 1 h' ( x ) =1 + = 0, x0 = 0. 2x h(0) = 0, h(4) = 6.∴ M = 6, m = 0. ( 4) f ( x ) = sin 2 x − x, | x |≤ π 2 f ' ( x ) = 2 cos 2 x −1 = 0, x0 = π 6 π 3π ππ π π f( )= − , f (− ) = , f ( ) = − . 6 26 2 2 2 2 π π ∴ M = ,m = − . 2 2 31 17. f ( x) = a ln x + bx 2 + x, f ' (1) = 0 a f ' ( x) = + 2bx + 1, x f ' (2) = 0 2 a + 2b +1 = 0 a = − 3 a ( ,∴ 2 + 4b +1 = 0 b = − 1 6 2 1 f ' ' ( x) = 2 − , f ' ' (1) > 0, f ' ' (2) < 0, 3x 3 ∴ f ( x1 )((((( f ( x2 )(((( 32 18. y = x − 1( x > 0), y ' (t ) = 2t , 2 y − (t − 1) = 2t ( x − t ) 1+ t 2 ( x = 0(( y = −t 2 − 1; ( y = 0(( x = . 2t 1 (1 + t 2 ) (1 + t 2 ) 2 (1) S (t ) = (1 + t 2 ) ⋅ = . 2 2t 4t 2 2 (1 + t )(3t − 1) 3 (2) S ' (t ) = = 0, t0 = . 2 4t 3 3 43 S min = S ( ) = . 3 9 (( (t , t − 1)(((((( 2 (3)( P(( 3 2 (− ( 3 3 2 33 19.((( x, ((( 512 .(( x 1024 f ( x) = x + x 1024 f ' ( x) = 1 − 2 = 0, x0 = 32. x f ' ' ( x0 ) > 0,∴((((((( 32(16(((((( 20.C (Q ) = 3 + Q, R (Q ) = 6Q − Q , 2 (1)(((( C ' (Q) = 1. (((( L' (Q ) = R ' (Q ) − C ' (Q) = 5 − 2Q, L' (2) = 1, L' (2.5) = 0, L' (3) = −1. (2)(((( 34 1p 21.Q = 160( ) , ( E (Q ) 3 p p 1p 1 ( .E (Q ) = ⋅ Q ' = ⋅160( ) ln 1p Q 3 3 160( ) 3 = −p ln 3. 22.E (Q ) = −2 p ln 2, ( E ( R ) p p ( .E ( R ) = ⋅ R ' (Q ) = ⋅ (Q + pQ' ) R (Q) pQ p = 1 + Q' = 1 + E (Q) = 1 − 2 p ln 2. Q 35 23.C (Q) = 5Q + 200, Q = 1000 −100 p.((((( Q ( . p = 10 − , (( 100 L(Q ) = pQ − C (Q ) Q2 =− + 5Q − 200. 100 Q L' (Q) = − + 5 = 0, Q0 = 250. 50 (((( Lmax = L(250) = 425. 36 ...
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