[tex]y=\sqrt{x-1}+6[/tex]
(Notice that we always have [tex]y\ge6[tex].)
Swap [tex]x[/tex] and [tex]y[/tex], then solve for [tex]y[/tex]:
[tex]x=\sqrt{y-1}+6[/tex]
[tex]x-6=\sqrt{y-1}[/tex]
[tex](x-6)^2=(\sqrt{y-1})^2[/tex]
[tex](x-6)^2=y-1[/tex]
[tex]y=(x-6)^2+1[/tex] (this is the inverse)
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This inverse is valid only for [tex]x\ge6[/tex]. Why?
Suppose we take [tex]x=0[/tex]. Then
[tex]y=(0-6)^2+1=37[/tex]
This would suggest that in the original equation, we should get [tex]x=37[/tex] when [tex]y=0[/tex]. But when we check this, we end up with
[tex]0=\sqrt{37-1}+6=\sqrt{36}+6=6+6=12[/tex]
which is clearly not true.
