f(x) = x^(1/3) + 1 so is f^-1(x) = (x-1)^(1/3) (?) thank you.
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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 $\displaystyle x^a$ is $\displaystyle x^{\tfrac 1a}$. here, $\displaystyle a=\tfrac 13$
Originally Posted by Moo Hello, Nope, but not far ^^ The inverse function of $\displaystyle x^a$ is $\displaystyle x^{\tfrac 1a}$. here, $\displaystyle a=\tfrac 13$ 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. $\displaystyle f(x) = \sqrt[3]{x} + 1$ $\displaystyle y = \sqrt[3]{x} + 1$ Switch x and y, then solve for y: $\displaystyle x = \sqrt[3]{y} + 1$ $\displaystyle \sqrt[3]{y} = x - 1$ $\displaystyle y = (x-1)^3$
... and how do we know that such function is invertible? What about the injection and surjection, hence bijection?
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