Question

SHOW YOUR WORK,

PLEASE HELP!!!!!!!!!

Answer 1

To find the inverse, we swap the variables y and x, then solve for the new y.

3a.
`y=(3)/(x-1)`

Swapping the variables:
`x=(3)/(y-1)`
Solving for y:
`x(y-1)=3 \\ y-1= (3)/(x) \\ y=1+(3)/(x)`
The domain of this inverse is
`x ≠ 0`.
3b.
`y=x^2-1`

Swapping:
`x = y^2 - 1`
Solving for y:
`y^2 = x + 1 \\ y = √(x+1)`
The domain of this inverse is
`x ≥ -1`.
3c.
`y=\sqrt[3]{(x-7)/(3)}`
Swapping:
`x=\sqrt[3]{(y-7)/(3)}`
Solving for y:
`x^3=(y-7)/(3) \\ y-7=3x^3 \\ y=3x^3+7`
The domain of this inverse is all real numbers.
4a.
`y=(3)/(x-1)`,
`y=1+(3)/(x)`

`y=(3)/((1+(3)/(x))-1) \\ y=(3)/(((3)/(x))) \\ y=x`

`y=1+(3)/(((3)/(x-1))) \\ y = 1+(x-1) \\ y = x`

4c.
`y=\sqrt[3]{(x-7)/(3)}`,
`y=3x^3+7`

`y=\sqrt[3]{((3x^3+7)-7)/(3)} \\ y=\sqrt[3]{(3x^3)/(3)} \\ y=\sqrt[3]{x^3} \\ y=x`

`y=3(\sqrt[3]{(x-7)/(3)})^3+7 \\ y = 3({(x-7)/(3)})+7 \\ y = (x-7)+7 \\ y=x`