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Question-59094




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
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}\:\mathrm{if}\:\mathrm{we}\:\mathrm{find}\:\mathrm{one}\:{f}\left({x}\right)\:\mathrm{for}\:\mathrm{which}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{not}\:\mathrm{true} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} \:\Rightarrow\:{f}''\left({x}\right)=\mathrm{6}{x} \\ $$$${f}\left({c}\right)−{f}\left({a}\right)\frac{{b}−{c}}{{b}−{a}}−{f}\left({b}\right)\frac{{c}−{a}}{{b}−{a}}−\frac{\mathrm{1}}{\mathrm{2}}\left({c}−{a}\right)\left({c}−{b}\right){f}''\left(\xi\right)=\mathrm{0} \\ $$$${c}^{\mathrm{3}} −{a}^{\mathrm{3}} \frac{{b}−{c}}{{b}−{a}}−{b}^{\mathrm{3}} \frac{{c}−{a}}{{b}−{a}}−\frac{\mathrm{1}}{\mathrm{2}}\left({c}−{a}\right)\left({c}−{b}\right)×\mathrm{6}\xi=\mathrm{0} \\ $$$$\frac{{c}^{\mathrm{3}} \left({b}−{a}\right)−{a}^{\mathrm{3}} \left({b}−{c}\right)−{b}^{\mathrm{3}} \left({c}−{a}\right)}{{b}−{a}}−\mathrm{3}\xi\left({c}−{a}\right)\left({c}−{b}\right)=\mathrm{0} \\ $$$$\frac{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\left({a}+{b}+{c}\right)}{\left({b}−{a}\right)}−\mathrm{3}\xi\left({c}−{a}\right)\left({c}−{b}\right)=\mathrm{0} \\ $$$$\left({a}−{c}\right)\left({b}−{c}\right)\left({a}+{b}+{c}\right)−\mathrm{3}\xi\left({a}−{c}\right)\left({b}−{c}\right)= \\ $$$$\left({a}−{c}\right)\left({b}−{c}\right)\left({a}+{b}+{c}−\mathrm{3}\xi\right)=\mathrm{0} \\ $$$$\Rightarrow\:{a}={c}\:\vee\:{b}={c}\:\vee\:\xi=\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$

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