Question

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

• A. $\frac{x+\sqrt{x^2-4}}{2}$
• B. $\frac{x}{1+x^2}$
• C. $\frac{x-\sqrt{x^2-4}}{2}$
• D. $x+\sqrt{x^2-4}$

Answer 1

Answer: (A)

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

Let $f\left(x\right)=x+\frac{1}{x}=y$
$\Rightarrow x=f^{-1}\left(y\right)$ & $x^2-yx+1=0$
Solving $x^2-yx+1$, we get
$x^2-yx+1=0$
$x=\frac{y\pm \sqrt{y^2-4}}{2}$
$\therefore f^{-1}=\frac{x+\sqrt{x^2-4}}{2}$
$\because$ $f$ is defined from $(1,\infty )\to (2,\infty )$ negative part is discarded.