Given:
The point (1,-1) lies on the given f(x).
Required:
We need to find the intersection of the inverse function.
Explanation:
The parent of the given function is
[tex]y=x^3[/tex]
The point (0,0) is moving to (1,-1)
[tex](x,y)=(0+1,0-1)[/tex]
The translation is 1 unit down and 1 unit right side.
The equation of the given graph is
[tex]y=(x-1)^3-1[/tex][tex]Replace\text{ }x=f^{-1}(y).[/tex][tex]y=(f^{-1}(y)-1)^3-1[/tex][tex]y+1=(f^{-1}(y)-1)^3-1+1[/tex][tex]y+1=(f^{-1}(y)-1)^3[/tex][tex]\sqrt[3]{y+1}=f^{-1}(y)-1[/tex][tex]\sqrt[3]{y+1}+1=f^{-1}(y)[/tex][tex]f^{-1}(x)=\sqrt[3]{x+1}+1[/tex]
The graph of the function f(x) and the inverse function:
Final answer:
The intersection point is (2.5, 2.5).
