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If $f:[1,\infty )\rightarrow\:[2,\infty)$ is defined as $f(x)=x+\large\frac{1}{x}$ then find $f^{-1}(x)$.

(A) $\large\frac{x+\sqrt {x^2-4}}{2}$
(B) $\large\frac{x-\sqrt {x^2-4}}{2}$
(C) $1+\sqrt{x^2-4}$
(D) $\large\frac{x}{1+x^2}$
Can you answer this question?
 
 

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Let $y=f(x)=x+\large\frac{1}{x}$
$\Rightarrow x=f^{-1}(y)$
$\Rightarrow\:xy=x^2+1$
$\Rightarrow\:x^2-xy+1=0$
$\Rightarrow\:x=\large\frac{y\pm \sqrt{y^2-4}}{2}=f^{-1}(y)$
$\Rightarrow\:f^{-1}(x)=\large\frac{x+\sqrt{x^2-4}}{2}$
since $\large\frac{x-\sqrt{x^2-4}}{2}$ is not in the range of $f(x)$.
answered May 27, 2013 by rvidyagovindarajan_1
 

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