1. ## inverse function

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

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

thank you.

2. 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, $\displaystyle a=\tfrac 13$

3. 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)

4. 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:
$\displaystyle x = \sqrt[3]{y} + 1$

$\displaystyle \sqrt[3]{y} = x - 1$

$\displaystyle y = (x-1)^3$

5. ... and how do we know that such function is invertible? What about the injection and surjection, hence bijection?