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Question Number 135215 by mnjuly1970 last updated on 11/Mar/21

$$.....calculuspreliminary....$$$$Q:f\left(x\right)=2^x-2^{-x}\Rightarrow f^{-1}\left(x\right)=???$$$$solution:$$$$y=2^x-2^{-x}.....$$$$y=\frac{2^{2x}-1}{2^x}\Rightarrow 2^{2x}-y{2}^x-1=0(\ast )...$$$$::2^x=t\Rightarrow t>0...✓....$$$$(\ast )\to ...t^2-ty-1=0....$$$$\mathrm{\Delta }=y^2+4>0...✓...$$$$t=\frac{y+\sqrt{y^2+4}}{2}......$$$$::2^x=\frac{y+\sqrt{y^2+4}}{2}\underset{bothsides}{\stackrel{takinglog}{\Rightarrow }}....$$$$x:={log}_2\left(\frac{y+\sqrt{y^2+4}}{2}\right)$$$$f^{-1}\left(x\right)={log}_2\left(\frac{x+\sqrt{x^2+4}}{2}\right)✓✓$$$$.........................$$$$$$$$$$

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