Este teorema es una continuidad del T. de Roller; manteniendo algunas consideraciones adicionales

Teorema del un valor medio

Este teorema es una continuidad del T. de Roller; manteniendo algunas consideraciones adicionales

. Que se utiliza en la formula de T.V.M los valores hallados de f(a) y (b)

. Se utiliza una formula especial, para el T.V.M.

. No siempre resultan 2 valores de c1 y c2

[f’(c) ] [b-a]= f(b) – f(a)

33. dada f(x)= -x2 + 8 x -6.. verificar si f(x) cumple con el lineamiento indicado en la aplicación de T.V.M.

f(x) = - x2+8 – 6

f (2) = - (2)2 + 8 (2) – 6 = - 4 + 16 – 6 = 6

f (3) = -  (3)2 + 8 (3) – 6 – 9 + 24 – 6 = 9

f (x) = - x2 + 8x – 6

f’ (x) = -  2x + 8

f’ (x) = 0  x=0

f’ (x) = -  2c + 8

[f ‘(c)] [b – a] = f (b) – f (a)

  [- 2c + 8] [3 – 2] = 9 – 6

  [- 2c + 8] [1] =3

    -2c + 8 =   - 2c = 3 – 8

    -2c= 5

C= 5/2 = 2.5

(0,2.5)

44. f(x) = x3 + x + 2… [2,4]

  f (a) = f (2) = 23 + 2 + 2 = 8 + 2 + 2 = 12

  f (b) = f(4) = 43 + 4 + 2 = 70

  f (x) = x3 + 3x + 2

 f’ (x) = 3x + 2 + 1

 f’ (x) = 0

 f’ (c) = 3c2 + 1

[f’(c) ] [b-a]= f(b) – f(a)

[3c2 + 1] [4 – 2] = 70 – 12

[3c2 + 1] [2] = 58

[6c2 + 2] =58

6c2 + 2 = 58

6c2 + = 58 – 2

6c2 = 56

C2 = 56/6

C= [pic 1]

C=± [pic 2]

C=± 3.055

45. f(x) = 2x3 - 3x2 + 3x - 1 … [-3,4]

  f (a) = f (-3) = 2(-3)3 – 3(-3)2 + 3(-3) – 1 = -91

  f (b) = f(4) = 2(4)3 – 3(4)2 + 3(4) – 1 = 91

  f (x) = 2x3 - 3x2 + 3x - 1

 f’ (x) = 6x2 – 6x + 3

 f’ (x) = 0         x = c

 f’ (c) = 6c2 – 6c + 3

[f’(c) ] [b-a]= f(b) – f(a)

[6c2 – 6c + 3] [4 – (-3) ] = 91 – (-91)

[6c2 – 6c + 3] [7] = 182

42c2 – 42c + 21 = 182

42c2 – 42c + 21 – 182 =0

42c2 – 42c - 161 = 0

a=42  b= -42  c= -161

c= [pic 3]

c= [pic 4]

c= [pic 5]

c= [pic 6]

c=     [pic 7]

 c=       c= [pic 8][pic 9]

46. f(x) = 4x3 - 3x2 + 6x - 2 … [-3,4]

  f (a) = f (-3) = 4(-3)3 – 3(-3)2 + 6(-6) – 2 = - 155

  f (b) = f(4) = 4(4)3 – 3(4)2 + 6(4) – 2 = 230

  f (x) = 4x3 - 3x2 + 6x - 2

 f’ (x) = 12x2 – 6x + 6

 f’ (x) = 0         x = c

 f’ (c) = 12c2 – 6c + 6

[f’(c) ] [b-a]= f(b) – f(a)

[12c2 – 6c + 6] [4 – (-3) ] = 230 – (-155)

[12c2 – 6c + 6] [7] = 385

84c2 – 42c + 42 = 385

84c2 – 42c + 42 - 385 = 0

84c2 – 42c -  343 = 0

a=84  b= -42  c= -343

c= [pic 10]

c= [pic 11]

c= [pic 12]

c= [pic 13]

c=     [pic 14]

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