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If $f:[1, \infty) \rightarrow[0, \infty)$ is given by $f(x)=x-\frac{1}{x}$, then $f^{-1}(x)=$
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Verified Answer
The correct answer is:
$\frac{1}{2}\left[x+\sqrt{x^2+4}\right]$
We have, $f(x)=x-\frac{1}{x}$
$$
\begin{aligned}
f\left(f^{-1}(x)\right) & =f^{-1}(x)-\frac{1}{f^{-1}(x)} \\
x & =\frac{\left(f^{-1} x\right)^2-1}{f^{-1}(x)}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow\left(f^{-1}(x)\right)^2-x f^{-1}(x)-1=0 \\
& \Rightarrow \quad f^{-1}(x)=\frac{x+\sqrt{x^2+4}}{2} \quad[\because x \in(1, \infty)] \\
& \therefore \quad f^{-1}(x)=\frac{1}{2}\left(x+\sqrt{x^2+4}\right)
\end{aligned}
$$
$$
\begin{aligned}
f\left(f^{-1}(x)\right) & =f^{-1}(x)-\frac{1}{f^{-1}(x)} \\
x & =\frac{\left(f^{-1} x\right)^2-1}{f^{-1}(x)}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow\left(f^{-1}(x)\right)^2-x f^{-1}(x)-1=0 \\
& \Rightarrow \quad f^{-1}(x)=\frac{x+\sqrt{x^2+4}}{2} \quad[\because x \in(1, \infty)] \\
& \therefore \quad f^{-1}(x)=\frac{1}{2}\left(x+\sqrt{x^2+4}\right)
\end{aligned}
$$
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