Notice that the graph of the function is a cubic polynomial. Also, the graph is moved one unit upwards, then, the function f(x) is:
[tex]f(x)=x^3+1[/tex]
now, we can see from the y and x intercepts, that if we evaluate x= 0 and x = 1, we get:
[tex]\begin{gathered} f(0)=-1 \\ f(1)=0 \end{gathered}[/tex]
then, applying the inverse function on both sides (we can do this since f(x) is a polynomial function and they always have inverse function), we get the following:
[tex]\begin{gathered} f^{-1}(f(0))=f^{-1}(-1) \\ \Rightarrow0=f^{-1}(-1) \end{gathered}[/tex]
we can see that the first point that is on the graph of the inverse function is (-1,0). Doing the same on the second equation, we get:
[tex]\begin{gathered} f^{-1}(f(1))=f^{-1}(0) \\ \Rightarrow f^{-1}(0)=1 \end{gathered}[/tex]
thus, the points that lie on the inverse function are (-1,0) and (0,1)
