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Re: Calculus Query
lets call f^-1(x) the inverse and (f(x))^-1 the recipricol
anyway
Surely if you have f(x) which has powers in, when you differentiate it the power will decrease by one (unless your using e^x)
reflecting it in y=x inverts the power
(eg f(x)=2x^2+3 f^-1(x) = (x-3)^(1/2) / 2
as such differentiating it only lowers the power by 1
so you can't use integers as when you lower the power it won't be afraction
if you want to try fractions, a/b - 1= b/a which the only thing that that has an equal distance from 1 on either side is 1/2 which would give 3/2 -1 (not equals) 2/3 so that can't work
so not fractions
i'm wondering maybe something hyperbolic or complex?
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