Answer
$ a.\quad f$ is one-to-one
$b.\quad f^{-1}(x)=x^{2}+1,\qquad x\geq 0$
Work Step by Step
$ a.\quad$
The graph of $f$ passes the horizontal line test (see below.)
(It is impossible to draw a horizontal line that intersects a function's graph more than once.)
It is one-to-one and has an inverse.
Note that $x\geq 1$ and $f(x)\geq 0.$.
The domain of $f^{-1}(x)$ is nonnegative numbers x.
$ b.\quad$
To find a formula for the inverse,
1. Replace $g(x)$ with $y.$
$y=\sqrt{x-1}, \qquad x\geq 1, y\geq 0$
2. Interchange $x$ and $y$. (This gives the inverse function.)
$x=\sqrt{y-1}, \qquad y\geq 1, x\geq 0$
3. Solve for $y.$
... square both sides,
$ x^{2}=y-1,\qquad y\geq 1, x\geq 0\qquad$ ... add 1
$x^{2}+1=y,\qquad y\geq 1, x\geq 0$
4. Replace $y$ with $f^{-1}(x)$ . (This is inverse function notation.)
$f^{-1}(x)=x^{2}+1,\qquad x\geq 0$
