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Question Number 182012 by CrispyXYZ last updated on 03/Dec/22

$$f\left(x\right)=9^x-m\cdot 3^x+m+6$$$$\mathrm{\exists }x\in \mathbb{R},f\left(x\right)+f(-x)=0$$$$\mathrm{find}\mathrm{the}\mathrm{range}\mathrm{of}m.$$

Answered by mr W last updated on 04/Dec/22

$$f\left(x\right)+f(-x)=0$$$$9^x+9^{-x}-m(3^x+3^{-x})+2(m+6)=0$$$${(3^x+3^{-x})}^2-m(3^x+3^{-x})+2(m+5)=0$$$$3^x+3^{-x}=\frac{m\pm \sqrt{{(m-4)}^2-56}}{2}⩾2$$$$\Rightarrow m\pm \sqrt{{(m-4)}^2-56}⩾4$$$$m-4⩾\pm \sqrt{{(m-4)}^2-56}$$$$\Rightarrow m-4⩾\sqrt{56}=2\sqrt{14}\Rightarrow m⩾4+2\sqrt{14}$$$$$$$$answerism\in [4+2\sqrt{14},+\mathrm{\infty })$$

Commented by manxsol last updated on 03/Dec/22

$$a+b\gg 2\sqrt{ab}$$$$3^x+3^{-x}\gg 2\sqrt{3^x.3^{-x}}=2$$

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