f(x) = x^(1/3) + 1

so is f^-1(x) = (x-1)^(1/3) (?)

thank you.

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- Aug 21st 2008, 07:55 AM2clientsinverse function
f(x) = x^(1/3) + 1

so is f^-1(x) = (x-1)^(1/3) (?)

thank you. - Aug 21st 2008, 07:57 AMMoo
- Aug 21st 2008, 08:09 AM2clients
- Aug 21st 2008, 08:10 AMChop Suey
- Aug 21st 2008, 11:16 AMKrizalid
... and how do we know that such function is invertible? What about the injection and surjection, hence bijection?