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If $f:[0, \infty) \rightarrow[2, \infty)$ is given by $f(x)=x+\frac{1}{x}$, then $f^{-1}(x)$ equals
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Verified Answer
The correct answer is:
$\frac{x+\sqrt{x^{2}-4}}{2}$
Let $f(x)=y=x+\frac{1}{x}$
or $x y=x^{2}+1$ or $x^{2}-x y+1=0$
Since, $x \in[0, \infty)$
$\therefore \quad D \geq 0 \Rightarrow y^{2}-4 \geq 0 \Rightarrow y \in[2, \infty)$
$\therefore \quad x=\frac{y \pm \sqrt{y^{2}-4}}{2}$
$\quad x=\frac{y-\sqrt{y^{2}-4}}{2}$ or $x=\frac{y-\sqrt{y^{2}-4}}{2}$
$\Rightarrow \quad f^{-1}(y)=\frac{y+\sqrt{y^{2}-4}}{2}$
or $f^{-1}(y)=\frac{y-\sqrt{y^{2}-4}}{2}$
Replace by $x$,
$f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}$
or
$f^{-1}(x)=\frac{x-\sqrt{x^{2}-4}}{2}$
or $x y=x^{2}+1$ or $x^{2}-x y+1=0$
Since, $x \in[0, \infty)$
$\therefore \quad D \geq 0 \Rightarrow y^{2}-4 \geq 0 \Rightarrow y \in[2, \infty)$
$\therefore \quad x=\frac{y \pm \sqrt{y^{2}-4}}{2}$
$\quad x=\frac{y-\sqrt{y^{2}-4}}{2}$ or $x=\frac{y-\sqrt{y^{2}-4}}{2}$
$\Rightarrow \quad f^{-1}(y)=\frac{y+\sqrt{y^{2}-4}}{2}$
or $f^{-1}(y)=\frac{y-\sqrt{y^{2}-4}}{2}$
Replace by $x$,
$f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}$
or
$f^{-1}(x)=\frac{x-\sqrt{x^{2}-4}}{2}$
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