Question Number 63485 by ANTARES VY last updated on 04/Jul/19 | ||
$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{F}}\left(\mathrm{2}\right)=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{F}}\left(−\mathrm{2}\right)=? \\ $$ | ||
Answered by MJS last updated on 04/Jul/19 | ||
$${f}\left({x}\right)={ax}+{b} \\ $$$${f}\left({x}−\mathrm{3}\right)+{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{2}{ax}−\mathrm{3}{a}+\mathrm{2}{b}=\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{2}{a}=\mathrm{2} \\ $$$$−\mathrm{3}{a}+\mathrm{2}{b}=−\mathrm{3} \\ $$$$\Rightarrow\:{a}=\mathrm{1}\wedge{b}=\mathrm{0} \\ $$$${f}\left({x}\right)={x} \\ $$$${F}\left({x}\right)=\int{xdx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$${F}\left(\mathrm{2}\right)=\mathrm{0}\:\Rightarrow\:{c}=−\mathrm{2} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2} \\ $$$${F}\left(−\mathrm{2}\right)=\mathrm{0} \\ $$ | ||