" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
f'(c)=(f(0)-f(-1))/(0--1)
f'(c)=(f(0)-f(-1))...(1)
f'(c)=(9c-2c^3)'
f'(c)=9-6c^2
f(0)=9*0-2*0^3
f(0)=0
f(-1)=9(-1)-2(-1)^3
f(-1)=-9-2(-1)
f(-1)=-9+2
f(-1)=-7
Reemplazando en (1):
9-6c^2=(0-(-7))
9-6c^2=(0+7)
9-6c^2=7
-6c^2=7-9
-6c^2=-2
-6c^2+2=0
c^2-2/6=0
c^2-1/3=0
(c+√(1/3))(c-√(1/3))=0
c=-√(1/3) y c=+√(1/3)
Como c ∈ <-1,0>
Entonces:
c=-√(1/3)