two more questions on my homework that I dont know how to do PLEASE HELP
one to one functions
attaching

Question

two more questions on my homework that I dont know how to do PLEASE HELP 
one to one functions
attaching

anonymous · Answer

attachment

anonymous · Answer

theres the first one

anonymous · Answer

heres the second

hero · Answer

The instructions are given on the assignment.

anonymous · Answer

yea but I tried solving for y and I got (x-2)^1/3  = wrong x^(1/3)-2 = wrong and other variations. I dont know what I'm doing wrong

hero · Answer

Show the work you did to arrive at that result, and I'll help you discover where you went wrong.

anonymous · Answer

(x-2)=cube root(y) then multiplied both sides by 1/3 got (x-2)^1/3=y

anonymous · Answer

the other one I did x=cube root(y)+2
x^(1/3)=y+2
x^(1/3)-2=y

hero · Answer

So you have

$y = \sqrt[3]{x} + 2$


You swapped x and y:
$x = \sqrt[3]{y} + 2$

Afterwards you subtracted 2 from both sides:
$x - 2 = \sqrt[3]{y}$

The next step is to CUBE both sides, rather multiply by 1/3

$(x - 2)^3 = y$

hero · Answer

And obviously y becomes $f^{-1}(x)$

anonymous · Answer

okay that makes sense

hero · Answer

You're welcome.