How do you find the inverse of # p(x)=x^3-3x^2+3x-1# and is it a function?

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Answer 1 · 2016-08-15T21:36:28

#f^(-1)(y) = root(3)(y)+1#

Let:

#y = p(x) = x^3-3x^2+3x-1=(x-1)^3#

Take the cube root of both ends to get:

#root(3)(y) = x-1#

Add #1# to both ends and transpose to find:

#x=root(3)(y)+1#

So:

#f^(-1)(y) = root(3)(y)+1#

which is a well defined function for any #y in RR#