inverse function

**2clients** · August 21st 2008, 07:55 AM

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

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

thank you.

**Moo** · August 21st 2008, 07:57 AM

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 $x^a$ is $x^{\tfrac 1a}$ . here, $a=\tfrac 13$

**2clients** · August 21st 2008, 08:09 AM

Originally Posted by Moo

Hello,

Nope, but not far ^^

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

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)

**Chop Suey** · August 21st 2008, 08:10 AM

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

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

$y = (x-1)^3$

**Krizalid** · August 21st 2008, 11:16 AM

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

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