If f:[1, infty) rightarrow [2, infty) is give | vedclass

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(A) Given $f(x) = x + \frac{1}{x}$.Let $y = f(x) = x + \frac{1}{x}$.Multiplying by $x$,we get $xy = x^2 + 1$,which implies $x^2 - xy + 1 = 0$.Using th

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$\frac{x + \sqrt{x^2 - 4}}{2}$

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(A) Given $f(x) = x + \frac{1}{x}$.Let $y = f(x) = x + \frac{1}{x}$.Multiplying by $x$,we get $xy = x^2 + 1$,which implies $x^2 - xy + 1 = 0$.Using the quadratic formula to solve for $x$:$x = \frac{y \pm \sqrt{y^2 - 4}}{2}$.Since the domain is $x \in [1, \infty)$,we must have $x \ge 1$.If we take $x = \frac{y - \sqrt{y^2 - 4}}{2}$,then for $y=2$,$x=1$,but for $y > 2$,this value would be less than $1$.Thus,we choose the positive root: $x = \frac{y + \sqrt{y^2 - 4}}{2}$.Therefore,$f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

If $f:[1, \infty) \rightarrow [2, \infty)$ is given by $f(x) = x + \frac{1}{x}$,then $f^{-1}(x)$ equals

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