#### Answer

(a) Because it passes the horizontal line test.
(b) The range is the segment $[-3,3]$ while the domain is the segment $[-1,3]$.
(c) $f^{-1}(2)=0$
(d) $f^{-1}(0)\approx1.7$

#### Work Step by Step

(a) Any horizontal line intercepts the graph of the function at most once. This means that every value of $y$ from the range corresponds to only one value of $x$ i.e. the function is one-to-one.
(b) From the graph we read that $x$ i.e. the argument takes values from $[-3,3]$ so this segment is the domain. On the other hand, the function takes values from $[-1,3]$ so this is the range. All of this is for $f$. Knowing that for $f^{-1}$ the domain and the range are reversed with respect to $f$we have that
Domain: $[-1,3]$
Range: $[-3,3]$
(c) We see that the graph passes through the point $(0,2)$ so $f(0)=2$, meaning that $f^{-1}(2)=0$.
(d) We see that the graph intercepts the $x$ around one third of a unit short of $x=-2$ which we estimate to be $x_0\approx1.7$. Because $f(x_0)=0$ we have that $f^{-1}(0)=x_0\approx1.7$.