Menu Close

Question-59094

Tinku Tara 0 Comments
Pin




Question Number 59094 by peter frank last updated on 04/May/19
Commented by Kunal12588 last updated on 04/May/19
I just want to ask from where do you get such question.
Answered by MJS last updated on 04/May/19
it′s not true if we find one f(x) for which it′s not true f(x)=x^3 ⇒ f′′(x)=6x f(c)−f(a)((b−c)/(b−a))−f(b)((c−a)/(b−a))−(1/2)(c−a)(c−b)f′′(ξ)=0 c^3 −a^3 ((b−c)/(b−a))−b^3 ((c−a)/(b−a))−(1/2)(c−a)(c−b)×6ξ=0 ((c^3 (b−a)−a^3 (b−c)−b^3 (c−a))/(b−a))−3ξ(c−a)(c−b)=0 (((a−b)(b−c)(c−a)(a+b+c))/((b−a)))−3ξ(c−a)(c−b)=0 (a−c)(b−c)(a+b+c)−3ξ(a−c)(b−c)= (a−c)(b−c)(a+b+c−3ξ)=0 ⇒ a=c ∨ b=c ∨ ξ=((a+b+c)/3) ⇒ it′s wrong
it′snottrueifwefindonef(x)forwhichit′snottruef(x)=x3⇒f″(x)=6xf(c)−f(a)b−cb−a−f(b)c−ab−a−12(c−a)(c−b)f″(ξ)=0c3−a3b−cb−a−b3c−ab−a−12(c−a)(c−b)×6ξ=0c3(b−a)−a3(b−c)−b3(c−a)b−a−3ξ(c−a)(c−b)=0(a−b)(b−c)(c−a)(a+b+c)(b−a)−3ξ(c−a)(c−b)=0(a−c)(b−c)(a+b+c)−3ξ(a−c)(b−c)=(a−c)(b−c)(a+b+c−3ξ)=0⇒a=c∨b=c∨ξ=a+b+c3⇒it′swrong

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *