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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

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

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