How do you find #f^-1(x)# given #f(x)=-2/(x-2)^3#?

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Answer 1 · 2016-12-29T00:18:45

Switch the x and y values.

#x = -2/(y -2)^3#

#x(y - 2)^3 = -2#

#(y - 2)^3 =-2/x#

#y - 2 = root(3)(-2/x)#

#y = root(3)(-2/x) + 2#

So, #f^-1(x) = root(3)(-2/x) + 2#.

Hopefully this helps!