Question

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

Answer 1

Step-1 Bijective function:

Given that $f:[1,\infty )\to [2,\infty )$ and $f\left(x\right)=x+\frac{1}{x}$

f is bijective,

For one-one

$$\begin{gathered} f\left(x_1\right)=f\left(x_2\right) \\ \Rightarrow x_1+\frac{1}{x_1}=x_2+\frac{1}{x_2} \\ \Rightarrow x_1-x_2=\frac{x_1-x_2}{x_1{x}_2} \\ \Rightarrow x_1=x_2 \end{gathered}$$

For onto

$x+\frac{1}{x}\ge 2\sqrt{x\times \frac{1}{x}}\ge 2$

So, range is equal to co-domain.

Step-2 Inverse function:

Therefore inverse exists.

$$\begin{gathered} y=f\left(x\right)=x+(1/x) \\ \Rightarrow x^2–xy+1=0 \\ \Rightarrow x=\frac{\{y\pm \sqrt{y^2-4}\}}{2} \\ \Rightarrow x=\frac{\{y+\sqrt{y^2-4}\}}{2},\frac{\{y–\sqrt{y^2-4}\}}{2} \end{gathered}$$

Since the values of $xandy$ are positive.

As we know that value of $x$ should be positive ,so $\frac{\{y-\sqrt{y^2-4}\}}{4}$ is not valid

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

Hence, the value of ${\mathit{f}}^{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is $\frac{\mathbf{\{}\mathbf{}\mathbf{x}\mathbf{+}\sqrt{{\mathbf{x}}^{\mathbf{2}}}\mathbf{–}\mathbf{4}\mathbf{}\mathbf{\}}}{\mathbf{2}}.$