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If \(f(x)=\log_{e}(x+\sqrt{1+x^{2}})\) then \(f^{-1}(x)\) is equal to |
\(\frac{e^{x}-e^{-x}}{2}\) \(\frac{e^{x}+e^{-x}}{2}\) \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\) \(\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\) |
\(\frac{e^{x}-e^{-x}}{2}\) |
\(y=\log_e(x+\sqrt{1+x^2})\) so $e^y=x+\sqrt{1+x^2}$ ...(1) as it is an odd function $e^{-y}-x+\sqrt{1+x^2}$ ...(2) so eq(1) - eq(2) $e^y-e^{-y}=2x$ so $\frac{e^y-e^{-y}}{2}$ so $f^{-1}(x)=\frac{e^x-e^{-x}}{2}$ |
