Question

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

Answer 1

theres the first one

Answer 2

heres the second

Answer 3

The instructions are given on the assignment.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

\((x - 2)^3 = y\)

Answer 9

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

Answer 10

okay that makes sense

Answer 11

You're welcome.