# inverse function

• Aug 21st 2008, 08:55 AM
2clients
inverse function
f(x) = x^(1/3) + 1

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

thank you.
• Aug 21st 2008, 08:57 AM
Moo
Hello,
Quote:

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 $x^a$ is $x^{\tfrac 1a}$. here, $a=\tfrac 13$ (Wink)
• Aug 21st 2008, 09:09 AM
2clients
Quote:

Originally Posted by Moo
Hello,

Nope, but not far ^^

The inverse function of $x^a$ is $x^{\tfrac 1a}$. here, $a=\tfrac 13$ (Wink)

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) (Worried)
• Aug 21st 2008, 09:10 AM
Chop Suey
Quote:

Originally Posted by 2clients
f(x) = x^(1/3) + 1

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

thank you.

$f(x) = \sqrt[3]{x} + 1$

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

Switch x and y, then solve for y:
$x = \sqrt[3]{y} + 1$

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

$y = (x-1)^3$
• Aug 21st 2008, 12:16 PM
Krizalid
... and how do we know that such function is invertible? What about the injection and surjection, hence bijection?