Question

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

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

Answer 1

Answer: (A)

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

$f\left(x\right)=x+\frac{1}{x}y=x+\frac{1}{x}xy=x^2+1x^2-xy+1=0$

Using quadratic formula

$x=\frac{y\pm \sqrt{y^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}x=\frac{y\pm \sqrt{y^2-4}}{2}$

$\Rightarrow x=\frac{y+\sqrt{y^2-4}}{2}\Rightarrow f^{-1}\left(x\right)=\frac{x+\sqrt{x^2-4}}{2}$