Question Number 182012 by CrispyXYZ last updated on 03/Dec/22
$${f}\left({x}\right)=\mathrm{9}^{{x}} −{m}\centerdot\mathrm{3}^{{x}} +{m}+\mathrm{6} \\ $$$$\exists{x}\in\mathbb{R},\:{f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\:\mathrm{range}\:\mathrm{of}\:{m}. \\ $$
Answered by mr W last updated on 04/Dec/22
$${f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{0} \\ $$$$\mathrm{9}^{{x}} +\mathrm{9}^{−{x}} −{m}\left(\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} \right)+\mathrm{2}\left({m}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} \right)^{\mathrm{2}} −{m}\left(\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} \right)+\mathrm{2}\left({m}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} =\frac{{m}\pm\sqrt{\left({m}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{56}}}{\mathrm{2}}\geqslant\mathrm{2} \\ $$$$\Rightarrow{m}\pm\sqrt{\left({m}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{56}}\geqslant\mathrm{4} \\ $$$${m}−\mathrm{4}\geqslant\pm\sqrt{\left({m}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{56}} \\ $$$$\Rightarrow{m}−\mathrm{4}\geqslant\sqrt{\mathrm{56}}=\mathrm{2}\sqrt{\mathrm{14}}\:\Rightarrow{m}\geqslant\mathrm{4}+\mathrm{2}\sqrt{\mathrm{14}} \\ $$$$ \\ $$$${answer}\:{is}\:{m}\in\left[\mathrm{4}+\mathrm{2}\sqrt{\mathrm{14}},+\infty\right) \\ $$
Commented by manxsol last updated on 03/Dec/22
$${a}+{b}\gg\mathrm{2}\sqrt{{ab}} \\ $$$$\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} \gg\mathrm{2}\sqrt{\mathrm{3}^{{x}} .\mathrm{3}^{−{x}} }=\mathrm{2} \\ $$