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Question Number 63485 by ANTARES VY last updated on 04/Jul/19

f(x−3)+f(x)=2x−3  F(2)=0.  F(−2)=?

$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{F}}\left(\mathrm{2}\right)=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{F}}\left(−\mathrm{2}\right)=? \\ $$

Answered by MJS last updated on 04/Jul/19

f(x)=ax+b  f(x−3)+f(x)=2x−3  2ax−3a+2b=2x−3  2a=2  −3a+2b=−3  ⇒ a=1∧b=0  f(x)=x  F(x)=∫xdx=(x^2 /2)+c  F(2)=0 ⇒ c=−2  F(x)=(x^2 /2)−2  F(−2)=0

$${f}\left({x}\right)={ax}+{b} \\ $$$${f}\left({x}−\mathrm{3}\right)+{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{2}{ax}−\mathrm{3}{a}+\mathrm{2}{b}=\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{2}{a}=\mathrm{2} \\ $$$$−\mathrm{3}{a}+\mathrm{2}{b}=−\mathrm{3} \\ $$$$\Rightarrow\:{a}=\mathrm{1}\wedge{b}=\mathrm{0} \\ $$$${f}\left({x}\right)={x} \\ $$$${F}\left({x}\right)=\int{xdx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$${F}\left(\mathrm{2}\right)=\mathrm{0}\:\Rightarrow\:{c}=−\mathrm{2} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2} \\ $$$${F}\left(−\mathrm{2}\right)=\mathrm{0} \\ $$

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