If f:[1, +infty) to [2, +infty) is given by f | vedclass

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(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$. Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equat

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

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(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$. Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equation $x^2 - yx + 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$. Since the domain of $f$ is $[1, +\infty)$,we must have $x \ge 1$. If we take the negative sign,$x = \frac{y - \sqrt{y^2 - 4}}{2}$. For $y \ge 2$,this value is $\le 1$. Specifically,as $y 	o \infty$,this expression approaches $0$. Thus,to satisfy $x \ge 1$,we must take the positive sign: $x = \frac{y + \sqrt{y^2 - 4}}{2}$. Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

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

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