Question

I need help finding the inverse of this problem

Answer 1

`y=√(x-1)+6`

(Notice that we always have
`y\ge6[tex].)</p><p>Swap [tex]x` and
`y`, then solve for
`y`:

`x=√(y-1)+6`

`x-6=√(y-1)`

`(x-6)^2=(√(y-1))^2`

`(x-6)^2=y-1`

`y=(x-6)^2+1` (this is the inverse)

###

This inverse is valid only for
`x\ge6`. Why?

Suppose we take
`x=0`. Then

`y=(0-6)^2+1=37`

This would suggest that in the original equation, we should get
`x=37` when
`y=0`. But when we check this, we end up with

`0=√(37-1)+6=√(36)+6=6+6=12`

which is clearly not true.