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