Este teorema es una continuidad del T. de Roller; manteniendo algunas consideraciones adicionales
Enviado por Helena • 11 de Diciembre de 2018 • 932 Palabras (4 Páginas) • 211 Visitas
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c= c= [pic 15][pic 16]
47. f(x) = 3x3 - 3x2 + 9x - 1 … [-3,4]
f (a) = f (-3) = 3(-3)3 – 3(-3)2 + 9(-3) – 1 = - 136
f (b) = f(4) = 3(4)3 – 3(4)2 + 9(4) – 1 = 179
f (x) = 3x3 - 3x2 + 9x - 1
f’ (x) = 9x2 – 6x + 9
f’ (x) = 0 x = c
f’ (c) = 9c2 – 6c + 9
[f’(c) ] [b-a]= f(b) – f(a)
[9c2 – 6c + 9] [4 – (-3) ] = 179 – (-136)
[9c2 – 6c + 9] [7] = 315
63c2 – 42c + 63 = 385
63c2 – 42c – 252 = 0
a=63 b= -42 c= -252
c= [pic 17]
c= [pic 18]
c= [pic 19]
c= [pic 20]
c= [pic 21]
c= c= [pic 22][pic 23]
48. f(x) = 2x3 - 3x2 + 6x - 3 … [-3,5]
f (a) = f (-3) = 2(-3)3 – 3(-3)2 + 6(-3) – 3 = - 102
f (b) = f(4) = 2(5)3 – 3(5)2 + 6(5) – 3 = 202
f (x) = 2x3 - 3x2 + 6x - 3
f’ (x) = 6x2 – 6x + 6
f’ (x) = 0 x = c
f’ (c) = 6c2 – 6c + 6
[f’(c) ] [b-a]= f(b) – f(a)
[ 6c2 – 6c + 6] [5 – (-3) ] = 202 – (-102)
[6c2 – 6c + 6] [7] = 304
48c2 – 48c + 48 = 304
48c2 – 48c + 48 - 304 = 0
48c2 – 48c - 256= 0
a=48 b= -48 c= -256
c= [pic 24]
c= [pic 25]
c= [pic 26]
c= [pic 27]
c= [pic 28]
c= c= [pic 29][pic 30]
48. f(x) = 2x3 - 3x2 + 6x - 3 … [-3,5]
f (a) = f (-3) = 2(-3)3 – 3(-3)2 + 6(-3) – 3 = - 102
f (b) = f(4) = 2(5)3 – 3(5)2 + 6(5) – 3 = 202
f (x) = 2x3 - 3x2 + 6x - 3
f’ (x) = 6x2 – 6x + 6
f’ (x) = 0 x = c
f’ (c) = 6c2 – 6c + 6
[f’(c) ] [b-a]= f(b) – f(a)
[ 6c2 – 6c + 6] [5 – (-3) ] = 202 – (-102)
[6c2 – 6c + 6] [7] = 304
48c2 – 48c + 48 = 304
48c2 – 48c + 48 - 304 = 0
48c2 – 48c - 256= 0
a=48 b= -48 c= -256
c= [pic 31]
c= [pic 32]
c= [pic 33]
c= [pic 34]
c= [pic 35]
c= c= [pic 36][pic 37]
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