Answer
See explanation
Work Step by Step
(a)
$$y=\sqrt{x+1}$$
Exchange $x$ and $y$:
$$x=\sqrt{y+1}$$
Solve for $y$:
$$x^2=y+1$$
$$y=x^2-1$$
Replace $y$ by $f^{-1}(x)$:
$$f^{-1}(x)=x^2-1,~\text{for}~x\geq0$$
(b) The graph of $f$ and $f^{-1}$ is as shown.
(c) $f^{-1}$ is a reflection of $f$ across the line $y=x$.
(d) The domain of $f$ is $[-1,\infty)$ and its range is $[0,\infty)$, while the domain of $f^{-1}$ is $[0,\infty)$ and its range is $[-1,\infty)$.
