f(x) = x^(1/3) + 1 so is f^-1(x) = (x-1)^(1/3) (?) thank you.
Follow Math Help Forum on Facebook and Google+
Hello, Originally Posted by 2clients f(x) = x^(1/3) + 1 is f^-1(x) = (x-1)^(1/3) thank you. Nope, but not far ^^ The inverse function of is . here,
Originally Posted by Moo Hello, Nope, but not far ^^ The inverse function of is . here, Thanks. I'm still unclear though: f(x) = x^(1/3) + 1 so y = x^(1/3) + 1 now if I solve for x why isn't that the inverse? x^(1/3) = y - 1 x = (y-1)^(1/3) f^-1(x) = (x-1)^(1/3)
Originally Posted by 2clients f(x) = x^(1/3) + 1 so is f^-1(x) = (x-1)^(1/3) (?) thank you. Switch x and y, then solve for y:
... and how do we know that such function is invertible? What about the injection and surjection, hence bijection?
View Tag Cloud