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f-x-9-x-m-3-x-m-6-x-R-f-x-f-x-0-find-the-range-of-m-




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

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