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