Question Number 193542 by SAMIRA last updated on 16/Jun/23
$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calcul}\:\:\:\:\mathrm{h}\left(\mathrm{X}\right)\:=\:\mathrm{f}\left(\mathrm{a}+\mathrm{X}\right)\:−\mathrm{b} \\ $$2) determine a and b such that h is odd
Commented by a.lgnaoui last updated on 17/Jun/23
$$\mathrm{odd}? \\ $$
Commented by SAMIRA last updated on 17/Jun/23
uneven
Answered by mr W last updated on 17/Jun/23
$$\left.\mathrm{2}\right) \\ $$$${h}\left({x}\right)=−{h}\left(−{x}\right) \\ $$$${f}\left({x}+{a}\right)−{b}=−{f}\left(−{x}+{a}\right)+{b} \\ $$$${f}\left({x}+{a}\right)=−{f}\left(−{x}+{a}\right)+\mathrm{2}{b} \\ $$$$\left({x}+{a}\right)^{\mathrm{3}} +\mathrm{3}\left({x}+{a}\right)^{\mathrm{2}} −\mathrm{1}=−\left(−{x}+{a}\right)^{\mathrm{3}} −\mathrm{3}\left(−{x}+{a}\right)^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{b} \\ $$$$\left({x}+{a}\right)^{\mathrm{3}} +\left(−{x}+{a}\right)^{\mathrm{3}} +\mathrm{3}\left({x}+{a}\right)^{\mathrm{2}} +\mathrm{3}\left(−{x}+{a}\right)^{\mathrm{2}} =\mathrm{2}+\mathrm{2}{b} \\ $$$$\mathrm{3}\left({a}+\mathrm{1}\right){x}^{\mathrm{2}} +{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} =\mathrm{1}+{b} \\ $$$$\Rightarrow{a}+\mathrm{1}=\mathrm{0}\:\Rightarrow{a}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}+{b}={a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} \:\Rightarrow{b}=\mathrm{1} \\ $$