Answer
$$f^{-1}(x)=1+\sqrt{x}$$
Work Step by Step
$$y=f(x)=x^2-2x+1\hspace{1cm}x\ge1$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y = x^2-2x+1$$ $$y=(x-1)^2$$
Since $x\ge1$, it follows that $(x-1)\ge0$. Therefore,
$$x-1=\sqrt{y}$$ $$x=1+\sqrt{y}$$
2) Interchange $x$ and $y$:
$$y=1+\sqrt{x}$$
Therefore, the inverse of function $y=f(x)=x^2-2x+1$, $x\ge1$ is the function $y=1+\sqrt{x}$. In other words, $$f^{-1}(x)=1+\sqrt{x}$$
