Question

Attaching question.
One to one function PLEASE HELP

Answer 1

Is that a cube?

\[\large f(x) = (x - 1)^3\]

Answer 2

yes

Answer 3

So to solve for the inverse...we merely switch the 'x' and the 'y' and then solve for 'y' again

\[\large y = (x - 1)^3\]

Lets switch the 'x' and 'y'

\[\large x = (y - 1)^3\]

And now we need to solve for 'y' again...how would we do that?

Answer 4

*Hint...I would take care of that cubed power first

Answer 5

Still there @brittlynn ?

Answer 6

sorry my wifi freaked out for a minute there

Answer 7

No problem ^_^

Answer 8

I'm not sure how to work with cubed powers

Answer 9

Okay...so we have

\[\large x = (y - 1)^3\]

So if we had a square there...we would take the square root of that to cancel it out right?

but here we have a cubed power...so we take a cubed root to cancel it out..

\[\large \sqrt[3]{x} = (y - 1)\]

did that make sense?

Answer 10

yes

Answer 11

then add 1?

Answer 12

Perfect...so now to just get 'y' by itself...we just need to add 1 to both sides of the equation

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

and that would be your \(\large f^{-1}(x)\)

Answer 13

ah okay, thank you!

Answer 14

of course! anytime :)