If f:[1, infty) rightarrow [0, infty) is give | vedclass

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(C) Given $f(x) = x - \frac{1}{x}$. Let $y = f(x)$,so $y = x - \frac{1}{x}$.To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:$y = \frac{x^2 - 1}{x

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$\frac{1}{2}\left[x+\sqrt{x^2+4}\right]$

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(C) Given $f(x) = x - \frac{1}{x}$. Let $y = f(x)$,so $y = x - \frac{1}{x}$.To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:$y = \frac{x^2 - 1}{x} \Rightarrow x^2 - yx - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{y \pm \sqrt{y^2 - 4(1)(-1)}}{2(1)} = \frac{y \pm \sqrt{y^2 + 4}}{2}$.Since the domain of $f$ is $[1, \infty)$,$x$ must be positive,so we take the positive root:$x = \frac{y + \sqrt{y^2 + 4}}{2}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x + \sqrt{x^2 + 4}}{2}$.

If $f:[1, \infty) \rightarrow [0, \infty)$ is given by $f(x) = x - \frac{1}{x}$,then $f^{-1}(x) =$

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