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

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Answer 1 · 2016-11-30T09:48:23

#The# inverse of #y = h(x)# is #x = y^(1/3)+1#. The graphs of both are the same.

#y = h(x)=(x-1)^3#. So, #x-1 = y^(1/3). And so, the inverse is

#x = h^(-1)(y) = y^(1/3)+1#.

Both graphs are the same. This is a verification for exactitude, using the Socratic

graphics facility.

graph{y-(x-1)^3=0 [-10, 10, -5, 5]}

graph{x-1-y^(1/3)=0 [-10, 10, -5, 5]}

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