Question 56078
As I explained in the previous exercise, I interchange the x and y, and solve for y.

y = 5x^3 + 8


Step 1:  Interchange the x and y:
x = 5y^3 + 8


Step 2:  Solve for y:
x-8 = 5y^3
{{{(x-8)/5 = y^3}}}


Take the cube root of each side:
{{{root(3, (x-8)/5) = y}}}

{{{root(3, (x/5)-(8/5)) = f^(-1)(x) }}}


This is what I got!!  I think you are correct!!  Why do you think it does not work???


R^2 at SCC