Step-by-step explanation:
f(x) = y = 64x³ - 1
y + 1 = 64x³
(y + 1)/64 = x³
[tex]x = \sqrt[3]{(y + 1) \div 64} [/tex]
[tex]x = \sqrt[3]{y + 1} \div 4[/tex]
that is the actual inverse function to calculate the original x value for a given y value of the original function.
but to make it a "normal" function, we need to rename the variables (x to y, y to x) :
[tex]y = \sqrt[3]{x + 1} \div 4[/tex]
this is now f^-1(x)
but careful, don't get confused, if dealing together with the original function, this "x" is actually standing for the "y" of the original function ...
